In projectile motion without air resistance, two launch angles that add up to
90° (called complementary angles) will produce the
same horizontal range when launched with the same initial speed.
the range for angles such as 30° and 60° will be identical. These are called
range pairs.
But what about the height?
Although complementary angles give the same range, they do not have the same:
maximum height
flight time
trajectory shape
A smaller angle (e.g., 30°) produces a lower, flatter path with a smaller height, while
a larger angle (e.g., 60°) produces a higher trajectory with a greater maximum height.
Thus, complementary angles give the same range, but not the same height or time of flight.
Problem: Arrow Shot Towards a Wall
An arrow is shot towards a wall 50m away with an initial velocity that makes an angle of 45° with the horizontal. It strikes the wall 35m above the ground. Assuming the arrow was released from ground level, find the initial velocity of the arrow.
Problem: Projectile Fired from the Top of a Building at an Angle
A projectile is fired with a muzzle velocity of 300m/s from a gun aimed upward at an angle of 20° with the horizontal, from the top of a building 30m. high above a level ground. With what velocity will it hit the ground, in m/s?
Problem: Bullet Shot Dead Center Toward a Target
A bullet shoots at 460 m/s at a target 130 m away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?
What is the maximum height achieved by the bullet?
Problem: Projectile Fired at a Distance from the Edge of a Cliff
50m from the edge of a cliff, a projectile is thrown at an angle of 30° with the horizontal. At what horizontal distance beyond the cliff does the ball strike the water if it just barely misses the edge of the cliff? The height of the cliff 62m.
Problem: Dropping a Load at a Horizontal Velocity
A small plane, flying at 180km/h at an altitude of 240m, is to drop an inflatable raft to flood victims stranded on a flat roof. How far from the roof should the pilot release the package so that it will land on the roof?
Problem: Ball Thrown just to Clear a Wall
A ball is thrown so that it just clears a 25-ft wall 100ft. away. If it left the hand 5 feet above the ground an at an angle of 60° to the horizontal, what was the initial velocity of the ball?
Problem: Projectile Thrown from the Edge of a Cliff
A projectile with a muzzle velocity of 550m/s is fired from a gun on top of a cliff 460m above the sea level. If the projectile hits the ocean surface 49.2 seconds after being fired, determine (a) the angle of elevation of the gun, (b) the range, and (c) the highest point attained above the cliff.
Problem: Concept of Complementary Angles Giving the Same Range
A catapult is placed 100 ft from the castle wall, which is 35 feet high. The soldier wants the burning bale of hay to clear the top of the wall and land 50 feet inside the castle wall. If the initial velocity of the bale is 70 feet per second, at what angle should the bale of hay be launched so that it travel 150 feet and pass over the castle wall. Use $g = 32 ft/sec^2$.
Exam Generator Problems
Additional board-style practice items for this topic.
A 45 cm Ć 45 cm square plate ABCD of uniform thickness is supported by three vertical strings. The first string is located at B, the second is located 15 cm from A along side AD, and the third is at the point midway of side CD. What percent of the load is carried by the first and third supports?
60%
40%
30%
70%
Place $A(0,0)$, $B(45,0)$, $C(45,45)$, $D(0,45)$; weight $W$ acts at the center $(22.5, 22.5)$. Supports: $R_1$ at $B(45,0)$, $R_2$ at $(0,15)$, $R_3$ at $(22.5,45)$. Moments about the y-axis: $45R_1 + 22.5R_3 = 22.5W \Rightarrow 2R_1 + R_3 = W$ Moments about the x-axis: $15R_2 + 45R_3 = 22.5W \Rightarrow R_2 + 3R_3 = 1.5W$ With $R_1 + R_2 + R_3 = W$, solving gives $R_3 = 0.4W$, $R_1 = 0.3W$. First + third $= 30\% + 40\%$ $\boxed{70\%}$
Neglecting air resistance, what minimum initial velocity must a rocket be fired to hit a target that is 5 km away?
186.2 kph
797.3 kph
670.3 kph
573.2 kph
Maximum range for a given speed occurs at $45^\circ$, where $R = \frac{v^2}{g}$. The minimum launch speed to reach range $R$ is: $v = \sqrt{Rg} = \sqrt{5000(9.81)} = 221.5$ m/s Convert to kph ($\times 3.6$): $\boxed{797.3 \text{ kph}}$
A projectile is fired 80 m above the level plane and reaches a horizontal distance of 650 m on that level plane. What is the initial velocity of the projectile if it is fired upward at an angle of 35° with the horizontal?
A block of mass M on a horizontal plane is being pushed by a constant horizontal force of 420 Newtons. The block has an initial velocity of 3.2 meters per second and a velocity of 15 meters per second after 3.6 seconds. The coefficients of static and kinetic frictions are 0.4 and 0.2, respectively.
Determine the mass of the block in kilograms.
80.2
75.6
90.3
84.9
What is the distance traveled by the block in meters?
54.8
32.8
43.3
38.7
What force is required to move the block by the same distance in 2 seconds?
1076 N
987 N
1213 N
1532 N
Part 1.
Acceleration from the velocity change: $a = \dfrac{15 - 3.2}{3.6} = 3.278$ m/sĀ². Newton's 2nd law with kinetic friction: $420 = m(a + \mu_k g) = m(3.278 + 0.2\times9.81)$: $$m = \frac{420}{5.24} = \boxed{80.2 \text{ kg}}$$
Part 2.
Over the 3.6 s with $v_0 = 3.2$ m/s and $a = 3.278$ m/sĀ²: $$s = v_0t + \tfrac{1}{2}at^2 = 3.2(3.6) + \tfrac{1}{2}(3.278)(3.6)^2 = \boxed{32.8 \text{ m}}$$
Part 3.
Cover the same 32.8 m in 2 s, keeping the initial velocity 3.2 m/s: $32.8 = 3.2(2) + \tfrac{1}{2}a'(2)^2 \Rightarrow a' = 13.18$ m/sĀ². $$F' = m(a' + \mu_k g) = 80.2(13.18 + 1.962) = \boxed{1213 \text{ N}}$$
Question Bank: t1818
MSTE - Dynamics / Engineering Mechanics / BEMz
Two identical sphere weighing 100 kN are each place in a container such that the lower sphere will be resting on a vertical wall and a horizontal wall and the other sphere will be resting on the lower sphere and a wall making an angle of 60 degrees with the horizontal. The line connecting the two centers of the spheres makes an angle of 30 degrees with the horizontal surface. Determine the reaction between the contact of the two spheres. Assume the walls to be frictionless.
150
120
180
100
For the upper sphere, the forces are its weight, the wall reaction, and the reaction from the lower sphere along the line of centers. The wall is at $60^\circ$, so its normal reaction is perpendicular to that wall. The line of centers is at $30^\circ$ to the horizontal. Horizontal equilibrium gives the wall reaction equal in magnitude to the sphere-contact reaction: $$R=N$$ Vertical equilibrium then gives $$N\sin30^\circ+R\sin150^\circ=100$$ Since $R=N$, $$0.5N+0.5N=100$$ $$N=100\text{ kN}$$ Therefore, the reaction between the spheres is $\boxed{100\text{ kN}}$.
Question Bank: t1819
MSTE - Dynamics / Engineering Mechanics / BEMz
The 5 m uniform steel beam has a mass of 600 kg and is to be lifted from the ring B with two chains, AB of length 3m, and CB of length 4m. Determine the tension T in chain AB when the beam is clear of the platform.
2.47 kN
3.68 kN
5.42 kN
4.52 kN
The chain lengths and beam length form a 3-4-5 triangle. For chain AB, the vertical component factor is $2.4/3=0.8$. For chain CB, it is $2.4/4=0.6$. The beam weight is $$W=600(9.81)=5.886\text{ kN}$$ Take moments about A to find the tension in CB: $$0.6T_{CB}(5)=W(2.5)$$ $$T_{CB}=4.905\text{ kN}$$ Vertical equilibrium: $$0.8T_{AB}+0.6T_{CB}=W$$ $$0.8T_{AB}+0.6(4.905)=5.886$$ $$T_{AB}=3.68\text{ kN}$$ Therefore, the tension in chain AB is $\boxed{3.68\text{ kN}}$.
Question Bank: t1821
MSTE - Dynamics / Engineering Mechanics / BEMz
Two men are just to lift a 300 kg weight of crowbar when the fulcrum for this lever is 0.3m from the weight and the man exerts their strengths at 0.9 m and 1.5 m respectively from the fulcrum. If the men interchange positions, they can raise a 340 kg weight. What force does each man exert?
25 kg, 40 kg
35 kg, 45 kg
40 kg, 50 kg
30 kg, 50 kg
Let the two men's forces be $A$ and $B$ in kgf. Taking moments about the fulcrum for the first position: $$0.9A+1.5B=300(0.3)=90$$ After they interchange positions: $$1.5A+0.9B=340(0.3)=102$$ Multiply by 10 and simplify: $$3A+5B=300$$ $$5A+3B=340$$ Solving gives $A=50$ and $B=30$ kgf. Thus, the two forces are $\boxed{30\text{ kg},\ 50\text{ kg}}$.
Question Bank: t1823
MSTE - Dynamics / Engineering Mechanics / BEMz
A simple beam having a span of 6m has a weight of 20 kN/m. It carries a concentrated load of 20 kN at the left end and 40 kN at 2m from the right end of the beam. If it is supported at 2m from the left end and the right end, compute the reaction at the right end of the beam.
40 kN
20 kN
50 kN
30 kN
Let the left end be $x=0$. The supports are at $x=2\text{ m}$ and $x=6\text{ m}$. The uniform load is $$20(6)=120\text{ kN}$$ acting at $x=3\text{ m}$. The 40 kN load is 2 m from the right end, so it is at $x=4\text{ m}$. Take moments about the support at $x=2\text{ m}$: $$R_R(4)+20(2)-120(1)-40(2)=0$$ $$4R_R+40-120-80=0$$ $$R_R=40\text{ kN}$$ Therefore, the reaction at the right end is $\boxed{40\text{ kN}}$.
Question Bank: t1824
MSTE - Dynamics / Engineering Mechanics / BEMz
When one boy is sitting 1.20 m from the center of a seesaw another boy must sit on the other side 1.50 m from the center to maintain an even balance. However, when the first boy carries an additional weight of 14 kg and sit 1.80 m from the center, the second boy must move 3m from the center to balance, Neglecting the weight of the see weight of the heaviest boy.
42 kg
35 kg
58 kg
29 kg
Let the first boy weigh $A$ kg and the second boy weigh $B$ kg. From the first balance condition, $$1.20A=1.50B$$ $$B=0.80A$$ With the first boy carrying an additional 14 kg: $$1.80(A+14)=3.00B$$ Substitute $B=0.80A$: $$1.80(A+14)=3.00(0.80A)$$ $$1.80A+25.2=2.40A$$ $$A=42\text{ kg}$$ Therefore, the keyed weight is $\boxed{42\text{ kg}}$.
Question Bank: t1825
MSTE - Dynamics / Engineering Mechanics / BEMz
A wire connects a middle of links AC and AB compute the tension in the wire if AC carries a uniform load of 600 N/m. AC is 4.5 m long and BC is 7.5 m. point A is hinged on the wall and joint C is also hinged connecting the links AC and CB. AC is horizontal while B is supported by roller acting on the wall AB.
2700 N
3600 N
300 N
2200 N
This linkage problem depends on the missing frame diagram and wire geometry. The load on member AC is $$wL=(600)(4.5)=2700\text{ N}$$ Using the source diagram/key for the link and wire equilibrium, the wire tension is $\boxed{2700\text{ N}}$.
Question Bank: t1827
MSTE - Dynamics / Engineering Mechanics / BEMz
A tripod whose legs are each 4 meters long supports a load of 1000 kg. the feet of the tripod are at the vertices of a horizontal equilateral triangle whose side are 3.5 meters. Determine the load of each leg.
386.19 kg
347.29 kg
214.69 kg
446.27 kg
By symmetry, each tripod leg carries the same axial force $T$. The horizontal distance from the center of an equilateral triangle to each foot is $$r=\frac{s}{\sqrt{3}}=\frac{3.5}{\sqrt{3}}=2.021\text{ m}$$ For a 4 m leg, the vertical height is $$h=\sqrt{4^2-r^2}=3.453\text{ m}$$ The vertical component factor is $h/4$, so $$3T\left(\frac{h}{4}\right)=1000$$ $$T=\frac{4000}{3(3.453)}=386.19\text{ kg}$$ Therefore, the load in each leg is $\boxed{386.19\text{ kg}}$.
Question Bank: t1828
MSTE - Dynamics / Engineering Mechanics / BEMz
A uniform square table top ABCD having sides 4m long is supported by three vertical supports at A, E and F, E is midway n the side BC and F is 1m from D along the side DC. Determine the share of load in percent carried by supports at A, E and F.
$A = 29%, E = 42%, F = 29%$
$A = 32%, E = 46%, F = 20%$
$A = 28%, E = 40%, F = 32%$
$A = 36%, E = 32%, F = 32%$
Use coordinates for the 4 m square: $A=(0,0)$, $E=(4,2)$, and $F=(1,4)$. The uniform tabletop weight acts at the centroid $(2,2)$. Let total weight be $W$. Force equilibrium: $$R_A+R_E+R_F=W$$ Moment balance using coordinates: $$4R_E+R_F=2W$$ $$2R_E+4R_F=2W$$ Solving, $$R_E=\frac{3}{7}W=42.9\%,\qquad R_F=\frac{2}{7}W=28.6\%$$ $$R_A=W-R_E-R_F=\frac{2}{7}W=28.6\%$$ Thus, $\boxed{A=29\%,\ E=42\%,\ F=29\%}$.
Question Bank: t1829
MSTE - Dynamics / Engineering Mechanics / BEMz
The square steel plate has a mass of 1800 kg with mass at its center G. Calculate the tension at each of the three cables with which the plate is lifted while remaining horizontal.
$Ta = Tb = 6.23 kN, Tc = 10.47 kN$
$Ta = Tb = 7.47 kN, Tc = 7.84 kN$
$Ta = Tb = 5.41 kN, Tc = 9.87 kN$
$Ta = Tb = 4.42 kN, Tc = 6.27 kN$
This lifting problem depends on the missing cable attachment geometry. The total plate weight is $$W=1800(9.81)=17.66\text{ kN}$$ The tensions must satisfy vertical force equilibrium and moment equilibrium about two horizontal axes so the square plate remains horizontal. Using the source diagram/key, the cable tensions are $\boxed{T_A=T_B=5.41\text{ kN},\ T_C=9.87\text{ kN}}$.
Question Bank: t1830
MSTE - Dynamics / Engineering Mechanics / BEMz
A horizontal Circular platform of radius R is supported at three points A, B and C on its circumference. A and B are 90 degrees apart and C is 120 degrees from A. The platform carries a vertical load of 400 kN at its center and 100 kN at a point d on the circumference diametrically opposite A. Compute the reaction at C.
253.45 kN
321.23 kN
310.10 kN
287.67 kN
Place the platform center at the origin and use radius $R$. Let A be at $(R,0)$, B at $(0,R)$, and C at $240^\circ$, so $C=(-0.5R,-0.866R)$. The 100 kN load is diametrically opposite A at $(-R,0)$. Force equilibrium: $$R_A+R_B+R_C=500$$ Moment balance in coordinate form gives $$R_A-0.5R_C=-100$$ $$R_B-0.866R_C=0$$ Thus $R_A=-100+0.5R_C$ and $R_B=0.866R_C$. Substitute into vertical equilibrium: $$-100+0.5R_C+0.866R_C+R_C=500$$ $$R_C=253.45\text{ kN}$$ Therefore, the reaction at C is $\boxed{253.45\text{ kN}}$.
Question Bank: t1831
MSTE - Dynamics / Engineering Mechanics / BEMz
A ladder 4m long having a mass of 15kg is resting against a floor and an wall for which the coefficients of static friction are 0.30 for the floor to which a man having a mass of 70 kg can climb without causing the plank to slip if the plank makes an angle of 40 degrees with the horizontal.
2
1
2.5
3
This ladder problem is missing part of the support-friction data and has a garbled statement. The usual solution is to draw the ladder free-body diagram, use limiting friction at the floor, and take moments about the lower end to solve for the man's distance along the ladder. Preserving the source key, the distance is $\boxed{1\text{ m}}$.
Question Bank: t1833
MSTE - Dynamics / Engineering Mechanics / BEMz
A uniform ladder on a wall at A and at the floor at B. Point A is 3.6m above the floor and point B is 1.5m away from the wall. Determine the minimum coefficient of friction at B required for a mass weighing 65 kg to use the ladder assuming that there is no friction at A.
0.42
0.50
0.48
0.54
With no friction at A, the wall reaction is horizontal. Taking the person at the top of the ladder for the limiting case, moment equilibrium about B gives $$R_A(3.6)=W(1.5)$$ $$\frac{R_A}{W}=\frac{1.5}{3.6}=0.4167$$ At B, the required friction is $F_B=R_A$ and the normal reaction is $N_B=W$, so $$\mu_B=\frac{F_B}{N_B}=0.4167$$ Therefore, the minimum coefficient of friction is $\boxed{0.42}$.
Question Bank: t1835
MSTE - Dynamics / Engineering Mechanics / BEMz
A 3.6 m ladder weighing 180 N is resting on a horizontal floor at A and on the wall at B making an angle of 30 degrees from the vertical wall. When a man weighing 800 N reaches a point 2.4m from the lower end (point A), the ladder is just about to slip. Determine the coefficient of friction between the ladder and the floor if the coefficient of friction between the ladder and the wall is 0.20.
0.35
0.42
0.28
0.56
The ladder makes $30^\circ$ with the vertical, so it makes $60^\circ$ with the floor. At the wall, $\mu_B=0.20$. Let $N_B$ be the wall normal reaction, so wall friction is $0.20N_B$ upward. Take moments about A. The top coordinates are $x=3.6\cos60^\circ=1.8\text{ m}$ and $y=3.6\sin60^\circ=3.118\text{ m}$. $$N_B(3.118)+0.20N_B(1.8)=180(0.9)+800(2.4\cos60^\circ)$$ $$N_B=322.6\text{ N}$$ Vertical equilibrium gives $$N_A+0.20N_B=180+800$$ $$N_A=915.5\text{ N}$$ Horizontal equilibrium gives floor friction $F_A=N_B=322.6\text{ N}$, so $$\mu_A=\frac{F_A}{N_A}=\frac{322.6}{915.5}=0.35$$ Therefore, the coefficient of friction at the floor is $\boxed{0.35}$.
Question Bank: t1837
MSTE - Dynamics / Engineering Mechanics / BEMz
Determine the distance āxā to which the 90 kg painter can climb without causing the 4m ladder to slip at its lower end A. The top of the 15 kg ladder has a small roller, and the ground coefficient of static friction is 0.25. the lower end of the ladder is 1.5 m away from the wall.
2.55 m
3.17 m
1.58 m
0.1 m
The top has a roller, so the wall reaction is horizontal only. The ladder height is $$h=\sqrt{4^2-1.5^2}=3.708\text{ m}$$ At impending slip at A, $$R_B=F_A=\mu N_A=0.25(90+15)g$$ Take moments about A. The ladder center is 0.75 m horizontally from A, and a painter at distance $x$ along the ladder is $0.375x$ horizontally from A. $$R_B(3.708)=15g(0.75)+90g(0.375x)$$ Canceling $g$: $$0.25(105)(3.708)=15(0.75)+90(0.375x)$$ $$x=2.55\text{ m}$$ Therefore, the painter can climb $\boxed{2.55\text{ m}}$ along the ladder.
Question Bank: t1838
MSTE - Dynamics / Engineering Mechanics / BEMz
The uniform pole of length 4m and mass 100kg is leaned against a vertical wall. If the coefficient of static friction between the supporting surfaces and the ends of the poles is 0.25, calculate the maximum angle Īø at which the pole may be placed with the vertical wall before it starts to slip.
28.07Ā°
26.57Ā°
31.6Ā°
33.5Ā°
For a uniform pole or ladder with equal friction coefficient at the floor and wall, the standard limiting condition gives the critical angle from the horizontal as $$\tan\alpha=\frac{1-\mu^2}{2\mu}$$ With $\mu=0.25$, $$\alpha=\tan^{-1}\left(\frac{1-0.25^2}{2(0.25)}\right)=61.93^\circ$$ So the angle with the vertical would be $$\theta=90^\circ-61.93^\circ=28.07^\circ$$ This is listed, but the source key selects $\boxed{33.5^\circ}$, indicating a likely figure or key mismatch.
Question Bank: t1848
MSTE - Dynamics / Friction / BEMz
A ladder is resting on a horizontal plane and a vertical wall. If the coefficient of friction between the ladder, the horizontal plane and the vertical wall is 0.40, determine the angle that the ladder makes with the horizontal at which it is about to slip.
46.4Ā°
33.6Ā°
53.13Ā°
64.13Ā°
For a uniform ladder with the same coefficient of friction at the floor and wall, the limiting angle satisfies $$\tan\theta=\frac{1-\mu^2}{2\mu}$$ With $\mu=0.40$: $$\tan\theta=\frac{1-0.40^2}{2(0.40)}=1.05$$ $$\theta=46.4^\circ$$ Therefore, the ladder is about to slip at $\boxed{46.4^\circ}$.
Question Bank: t1857
MSTE - Dynamics / Friction / BEMz
Find the value of P acting to the left that is required to pull the wedge out under the 500kg block. Angle of friction is 20Ā° for all contact surfaces.
253.80 kg
242.49 kg
432.20 kg
120.50 kg
This wedge problem depends on the missing figure, including the wedge angle and contact layout. The solution should use the angle of friction $\phi=20^\circ$ at each contact and draw the force polygons or free-body diagrams for the block and wedge while the wedge is pulled out. Using the source diagram/key, the required pull is $\boxed{253.80\text{ kg}}$.
Question Bank: t1895
MSTE - Dynamics / Projectile Motion / BEMz
A ball is shot at a ground level at an angle of 60 degrees with the horizontal with an initial velocity of 100 m/s. Determine the height of the ball after 2 seconds.
162.46 m
153.59 m
175.48 m
186.42 m
Use vertical projectile displacement from ground level. $$y=v_0\sin\theta\,t-\frac{1}{2}gt^2$$ $$y=(100\sin60^\circ)(2)-\frac{1}{2}(9.81)(2^2)$$ $$y=173.21-19.62=153.59\text{ m}$$ Therefore, the height after 2 seconds is $\boxed{153.59\text{ m}}$.
Question Bank: t1896
MSTE - Dynamics / Projectile Motion / BEMz
A ball is shot at an average speed of 200 m/s at an angle of 20Ā° with the horizontal. What would be the velocity of the ball after 8 seconds?
188.21 m/s
154.34 m/s
215.53 m/s
198.37 m/s
Resolve the velocity into components. $$v_x=200\cos20^\circ=187.94\text{ m/s}$$ $$v_y=200\sin20^\circ-9.81(8)=-10.08\text{ m/s}$$ The speed after 8 seconds is $$v=\sqrt{v_x^2+v_y^2}=\sqrt{187.94^2+(-10.08)^2}=188.21\text{ m/s}$$ Thus, the velocity magnitude is $\boxed{188.21\text{ m/s}}$.
Question Bank: t1897
MSTE - Dynamics / Projectile Motion / BEMz
A projectile has a velocity of 200 m/s acting at an angle 20 degrees with the horizontal. How long will it take for the projectile to hit the ground surface?
$13.95 \sec$
$15.75 \sec$
$10.11 \sec$
$24.23 \sec$
For a projectile returning to the same level, $$T=\frac{2v_0\sin\theta}{g}$$ Substitute the given values: $$T=\frac{2(200)\sin20^\circ}{9.81}=13.95\text{ s}$$ Therefore, the projectile hits the ground after $\boxed{13.95\text{ s}}$.
