Practice important
Impact Angle
MCQ questions and answers.
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10
MCQ Questions
1. A projectile is fired horizontally with speed 20 m/s from the top of a 20 m high cliff. Taking g = 10 m/s^2, what angle does its velocity make with the horizontal just before impact?
A.
30° below the horizontal
B.
45° below the horizontal
C.
60° below the horizontal
D.
75° below the horizontal
Correct Answer:
45° below the horizontal
Explanation:
The horizontal velocity remains 20 m/s. The time of fall is sqrt(2h/g) = sqrt(40/10) = 2 s, so the downward vertical velocity at impact is gt = 20 m/s. Hence tan(theta) = 20/20 = 1, giving theta = 45° below the horizontal.
2. A projectile is launched from ground level with speed 25 m/s at an angle whose sine and cosine are 3/5 and 4/5, respectively. It strikes a vertical wall 10 m above the point of projection. Taking g = 10 m/s^2, what is the angle of impact with the horizontal on the descending path?
A.
tan^-1(1/2) below the horizontal
B.
tan^-1(1/3) below the horizontal
C.
tan^-1(1/4) below the horizontal
D.
tan^-1(1/5) below the horizontal
Correct Answer:
tan^-1(1/4) below the horizontal
Explanation:
The horizontal and initial vertical components are 20 m/s and 15 m/s. At height 10 m, 10 = 15t - 5t^2, giving t = 2 s on the descending path. Thus the vertical velocity is 15 - 10(2) = -5 m/s. Therefore tan(theta) = 5/20 = 1/4, so the impact angle is tan^-1(1/4) below the horizontal.
3. A projectile is launched horizontally from a height of 45 m and strikes the ground at an angle of 60° below the horizontal. Taking g = 10 m/s^2, what was its horizontal launch speed?
A.
5sqrt(3) m/s
B.
10 m/s
C.
10sqrt(3) m/s
D.
20sqrt(3) m/s
Correct Answer:
10sqrt(3) m/s
Explanation:
The fall time is t = sqrt(2h/g) = sqrt(90/10) = 3 s. The vertical impact speed is gt = 30 m/s. Since tan 60° = vertical speed/horizontal speed, the horizontal speed is 30/sqrt(3) = 10sqrt(3) m/s.
4. A projectile is launched with speed 20 m/s at 30° above the horizontal from a point and lands 5 m below the point of projection. Taking g = 10 m/s^2, what is the tangent of its impact angle below the horizontal?
A.
(sqrt(2) + 1)/sqrt(3)
B.
(sqrt(2) - 1)/sqrt(3)
C.
1/sqrt(2)
D.
sqrt(3)/(sqrt(2) - 1)
Correct Answer:
(sqrt(2) - 1)/sqrt(3)
Explanation:
The initial components are u_x = 10sqrt(3) m/s and u_y = 10 m/s. Using -5 = 10t - 5t^2 gives t = 1 + sqrt(2) s. The final vertical velocity is 10 - 10t = -10sqrt(2) m/s. Thus tan(theta) = 10(sqrt(2)-1)/(10sqrt(3)) = (sqrt(2)-1)/sqrt(3).
5. A projectile has horizontal velocity 10 m/s and initial vertical velocity 20 m/s. It strikes a target 30 m horizontally from the launch point. Taking g = 10 m/s^2, what is the impact angle?
A.
30° below the horizontal
B.
45° below the horizontal
C.
60° below the horizontal
D.
75° below the horizontal
Correct Answer:
45° below the horizontal
Explanation:
The time to reach the target is t = 30/10 = 3 s. Its vertical velocity then is v_y = 20 - 10(3) = -10 m/s, while the horizontal velocity remains 10 m/s. Therefore tan(theta) = 10/10 = 1, so the impact angle is 45° below the horizontal.
6. A projectile is launched from ground level and returns to the same level with an impact angle of 37° below the horizontal. If its speed at impact is 25 m/s, what was its initial vertical component of velocity? Use sin 37° = 3/5.
A.
12 m/s
B.
15 m/s
C.
20 m/s
D.
24 m/s
Correct Answer:
15 m/s
Explanation:
At impact, the downward vertical component is 25 sin 37° = 25(3/5) = 15 m/s. For a projectile returning to its original height, the vertical component at impact has the same magnitude as the initial vertical component. Hence the initial vertical component was 15 m/s.
7. Two projectiles are fired from the same point with the same speed and at angles 25° and 65° above the horizontal. Both return to the original level. What is the difference between their impact angles, measured below the horizontal?
A.
25°
B.
40°
C.
65°
D.
90°
Correct Answer:
40°
Explanation:
For a projectile landing at the same level from which it was launched, the impact angle below the horizontal equals the launch angle above the horizontal. Thus the two impact angles are 25° and 65°, whose difference is 40°.
8. A projectile remains in flight for 4 s and travels 80 m horizontally before landing at its launch level. If its impact angle is 45° below the horizontal, what was its launch speed?
A.
20 m/s
B.
20sqrt(2) m/s
C.
40 m/s
D.
40sqrt(2) m/s
Correct Answer:
20sqrt(2) m/s
Explanation:
The horizontal component is u_x = 80/4 = 20 m/s. At impact, the vertical component is v_y = -gT/2 = -20 m/s because the projectile returns to its launch level. The impact angle is therefore 45°, consistent with equal component magnitudes. The launch vertical component is +20 m/s, so the launch speed is sqrt(20^2 + 20^2) = 20sqrt(2) m/s.
9. A projectile is launched with horizontal and vertical velocity components 24 m/s and 18 m/s, respectively, from the top of a 36 m high cliff. Taking g = 10 m/s^2, it hits the ground after 4 s. What is its impact angle below the horizontal?
A.
tan^-1(1/2) below the horizontal
B.
tan^-1(5/12) below the horizontal
C.
tan^-1(7/12) below the horizontal
D.
tan^-1(2) below the horizontal
Correct Answer:
tan^-1(7/12) below the horizontal
Explanation:
The horizontal velocity remains 24 m/s. After 4 s, the vertical velocity is v_y = 18 - 10(4) = -22 m/s. Therefore tan(theta) = 22/24 = 11/12, not 7/12. Since the stated 4 s is inconsistent with the given height, the physically valid descending time must be obtained from 0 = 36 + 18t - 5t^2, giving t = 4.8 s. Then v_y = -30 m/s, so tan(theta) = 30/24 = 5/4. Thus none of the listed options is correct.
10. A projectile is launched with speed 20 m/s at 37° above the horizontal and lands 15 m below its launch point. Take sin 37° = 3/5, cos 37° = 4/5, and g = 10 m/s^2. What is the tangent of its impact angle below the horizontal?
A.
1/2
B.
3/4
C.
1
D.
5/4
Correct Answer:
5/4
Explanation:
The components are u_x = 16 m/s and u_y = 12 m/s. The vertical displacement equation is -15 = 12t - 5t^2, giving t = 3 s on the descending path. Thus the vertical impact velocity is 12 - 30 = -18 m/s. Therefore tan(theta) = 18/16 = 9/8, so the correct value should be 9/8; the provided options are inconsistent.
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