physics : class11 Time of Flight : Projectile Motion : Kinematics MCQ Questions and Answers
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Time of Flight
MCQ questions and answers.
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MCQ Questions
1. A projectile is launched with speed 20 m/s at an angle of 30° to the horizontal. Taking g = 10 m/s², what is its time of flight when it lands at the same level?
A.
1 s
B.
2 s
C.
3 s
D.
4 s
Correct Answer:
2 s
Explanation:
For a projectile returning to the same level, time of flight T = 2u sinθ/g. Thus, T = 2 × 20 × sin30°/10 = 2 s.
2. A projectile is thrown vertically upward with an initial speed of 30 m/s. Taking g = 10 m/s², how long does it remain in the air before returning to the point of projection?
A.
3 s
B.
4 s
C.
6 s
D.
9 s
Correct Answer:
6 s
Explanation:
For vertical projection, the total time of flight is T = 2u/g. Therefore, T = 2 × 30/10 = 6 s.
3. For a projectile launched and landing at the same level, which expression gives its time of flight?
A.
u cosθ/g
B.
u sinθ/g
C.
2u sinθ/g
D.
2u cosθ/g
Correct Answer:
2u sinθ/g
Explanation:
The vertical component of the initial velocity is u sinθ. Since the projectile returns to the same level, its time of flight is T = 2u sinθ/g.
4. Two projectiles are launched with the same speed at angles 30° and 60° to the horizontal. Which projectile has the greater time of flight when both land at the same level?
A.
The projectile launched at 30°
B.
The projectile launched at 60°
C.
Both have the same time of flight
D.
The time of flight cannot be compared
Correct Answer:
The projectile launched at 60°
Explanation:
Time of flight is proportional to sinθ. Since sin60° is greater than sin30°, the projectile launched at 60° remains in the air longer.
5. A projectile is launched at an angle of 90° to the horizontal. If its initial speed is 20 m/s and g = 10 m/s², what is its time of flight?
A.
1 s
B.
2 s
C.
4 s
D.
20 s
Correct Answer:
4 s
Explanation:
For θ = 90°, the projectile is projected vertically upward. Its time of flight is T = 2u/g = 2 × 20/10 = 4 s.
6. A projectile has an initial vertical component of velocity equal to 15 m/s and lands at the same level from which it was launched. Taking g = 10 m/s², what is its time of flight?
A.
1.5 s
B.
2 s
C.
3 s
D.
15 s
Correct Answer:
3 s
Explanation:
When the projectile returns to its initial level, time of flight is T = 2u_y/g, where u_y is the initial vertical component. Thus, T = 2 × 15/10 = 3 s.
7. For a projectile returning to the same height, its time of flight is doubled while the angle of projection remains unchanged. What happens to its initial speed?
A.
It becomes half
B.
It remains unchanged
C.
It becomes double
D.
It becomes four times
Correct Answer:
It becomes double
Explanation:
Since T = 2u sinθ/g, for a fixed angle and fixed gravitational acceleration, time of flight is directly proportional to initial speed. Therefore, doubling T doubles u.
8. A projectile is launched with speed 10 m/s at an angle of 90° to the horizontal. Taking g = 10 m/s², what is the time taken to return to its point of projection?
A.
0.5 s
B.
1 s
C.
2 s
D.
10 s
Correct Answer:
2 s
Explanation:
For vertical projection, T = 2u/g. Substituting u = 10 m/s and g = 10 m/s² gives T = 2 s.
9. A projectile is launched at 30° and another at 90°, both with the same initial speed and landing at the same level. Which statement about their time of flight is correct?
A.
The 30° projectile has the greater time of flight
B.
The 90° projectile has the greater time of flight
C.
Both have zero time of flight
D.
Both have the same time of flight
Correct Answer:
The 90° projectile has the greater time of flight
Explanation:
Time of flight is T = 2u sinθ/g. For the same speed, sin90° is greater than sin30°, so the projectile launched at 90° has the greater time of flight.
10. A projectile is launched with speed 25 m/s at an angle of 53° to the horizontal. Taking sin53° = 0.8 and g = 10 m/s², what is its time of flight when it lands at the same level?
A.
2 s
B.
3 s
C.
4 s
D.
5 s
Correct Answer:
4 s
Explanation:
Using T = 2u sinθ/g, T = 2 × 25 × 0.8/10 = 4 s.