Projectile Motion — NEET Physics MCQs with Solutions
Free NEET Physics Projectile Motion MCQs with step-by-step solutions (20 questions). Part of Motion in 2D. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Projectile Motion · easy · numerical
A ball is projected with a speed of 20 m/s at 30° above the horizontal. Its time of flight is (g = 10 m/s²):
A. 1 s
B. 2 s ✓ Correct
C. 4 s
D. 2√3 ≈ 3.5 s
Solution: T = 2u sinθ/g = (2 × 20 × 0.5)/10 = 2 s.
Q2 — Projectile Motion · easy · numerical
A football is kicked with a speed of 40 m/s at 30° above the horizontal. The maximum height reached is (g = 10 m/s²):
A. 60 m
B. 40 m
C. 20 m ✓ Correct
D. 80 m
Solution: H = u²sin²θ/(2g) = (1600 × 0.25)/20 = 20 m.
Q3 — Projectile Motion · easy · numerical
A ball is projected at 20 m/s at an angle of 45° with the ground. Its horizontal range is (g = 10 m/s²):
A. 20 m
B. 10 m
C. 40 m ✓ Correct
D. 80 m
Solution: R = u²sin2θ/g = (400 × sin90°)/10 = 40 m.
Q4 — Projectile Motion · easy · numerical
A cricket ball can be thrown with a maximum speed of 30 m/s. The greatest horizontal distance it can cover is (g = 10 m/s²):
A. 45 m
B. 30 m
C. 90 m ✓ Correct
D. 180 m
Solution: Maximum range occurs at 45°: R_max = u²/g = 900/10 = 90 m.
Q5 — Projectile Motion · easy · numerical
A projectile is fired at 50 m/s at 60° above the horizontal. Its speed at the highest point is:
A. 25 m/s ✓ Correct
B. 50 m/s
C. 0 m/s
D. 25√3 ≈ 43.3 m/s
Solution: At the top only the horizontal component remains: v = u cos60° = 50 × 0.5 = 25 m/s.
Q6 — Projectile Motion · easy · numerical
A ball is thrown horizontally at 10 m/s from the top of a 45 m tower. The time it takes to reach the ground is (g = 10 m/s²):
A. 3 s ✓ Correct
B. 4.5 s
C. 2 s
D. 9 s
Solution: t = √(2h/g) = √(2 × 45/10) = √9 = 3 s (independent of the horizontal speed).
Q7 — Projectile Motion · easy · numerical
A stone is thrown horizontally at 12 m/s from the top of a 20 m building. The horizontal distance from the base at which it lands is (g = 10 m/s²):
A. 40 m
B. 24 m ✓ Correct
C. 12 m
D. 48 m
Solution: t = √(2 × 20/10) = 2 s, so R = 12 × 2 = 24 m.
Q8 — Projectile Motion · easy · numerical
A plane flying horizontally at 100 m/s at a height of 500 m releases a bomb. The bomb strikes the ground at a horizontal distance from the release point of (g = 10 m/s²):
A. 2000 m
B. 1000 m ✓ Correct
C. 500 m
D. 100 m
Solution: t = √(2 × 500/10) = 10 s; the bomb keeps the plane speed, so x = 100 × 10 = 1000 m.
Q9 — Projectile Motion · easy · numerical
A ball projected at 30° above the horizontal rises to a maximum height of 5 m. Its speed of projection is (g = 10 m/s²):
A. 10 m/s
B. 10√2 ≈ 14.1 m/s
C. 40 m/s
D. 20 m/s ✓ Correct
Solution: H = u²sin²30°/(2g): 5 = u² × 0.25/20, so u² = 400 and u = 20 m/s.
Q10 — Projectile Motion · easy · numerical
The horizontal range of a football is 4 times its maximum height. The angle of projection is:
A. 60°
B. 30°
C. 45° ✓ Correct
D. 75°
Solution: tanθ = 4H/R; with R = 4H, tanθ = 1, so θ = 45°.
Q11 — Projectile Motion · easy · numerical
Two balls are projected with the same speed, one at 30° and the other at 60° above the horizontal. The ratio of their maximum heights (30° to 60°) is:
A. 1 : 3 ✓ Correct
B. 1 : 2
C. 1 : √3
D. 3 : 1
Solution: H ∝ sin²θ, so the ratio is sin²30° : sin²60° = (1/4) : (3/4) = 1 : 3.
Q12 — Projectile Motion · easy · numerical
A ball thrown at 20 m/s at 30° above the horizontal has a certain range. With the same speed, the other angle of projection giving the same range is:
A. 90°
B. 60° ✓ Correct
C. 75°
D. 45°
Solution: Ranges are equal at complementary angles: 90° − 30° = 60°.
Q13 — Projectile Motion · medium · numerical
A ball is projected with velocity (12î + 15ĵ) m/s, where ĵ is vertically up. Its speed 1 s after projection is (g = 10 m/s²):
A. 5 m/s
B. 17 m/s
C. 13 m/s ✓ Correct
D. 12 m/s
Solution: vₓ = 12 m/s stays constant; v_y = 15 − 10 × 1 = 5 m/s, so v = √(12² + 5²) = 13 m/s.
Q14 — Projectile Motion · medium · numerical
A ball is projected with a vertical velocity component of 20 m/s. The times at which it is at a height of 15 m are (g = 10 m/s²):
A. 0.5 s and 3.5 s
B. 1 s and 3 s ✓ Correct
C. 2 s and 4 s
D. 1 s and 2 s
Solution: 15 = 20t − 5t² gives t² − 4t + 3 = 0, so t = 1 s and t = 3 s.
Q15 — Projectile Motion · medium · numerical
For a projectile, the maximum height equals (√3/4) times the horizontal range. The angle of projection is:
A. 30°
B. 75°
C. 45°
D. 60° ✓ Correct
Solution: tanθ = 4H/R = 4 × (√3/4) = √3, so θ = 60°.
Q16 — Projectile Motion · medium · numerical
A projectile launched at 45° has a horizontal range of 250 m. Its speed of projection is (g = 10 m/s²):
A. 25 m/s
B. 50 m/s ✓ Correct
C. 25√2 ≈ 35.4 m/s
D. 100 m/s
Solution: At 45°, R = u²/g, so u = √(gR) = √(10 × 250) = 50 m/s.
Q17 — Projectile Motion · medium · numerical
The time of flight of a projectile is 4 s. Its maximum height is (g = 10 m/s²):
A. 10 m
B. 80 m
C. 20 m ✓ Correct
D. 40 m
Solution: u_y = gT/2 = 10 × 4/2 = 20 m/s; H = u_y²/(2g) = 400/20 = 20 m.
Q18 — Projectile Motion · medium · numerical
A ball thrown horizontally from the top of an 80 m tower lands 120 m from its base. The speed with which it was thrown is (g = 10 m/s²):
A. 15 m/s
B. 30 m/s ✓ Correct
C. 40 m/s
D. 60 m/s
Solution: t = √(2 × 80/10) = 4 s, so u = 120/4 = 30 m/s.
Q19 — Projectile Motion · medium · numerical
A football is kicked horizontally at 24 m/s off a 5 m high ledge. Its speed just before hitting the ground is (g = 10 m/s²):
A. 10 m/s
B. 26 m/s ✓ Correct
C. 34 m/s
D. 24 m/s
Solution: t = √(2 × 5/10) = 1 s, so v_y = 10 m/s; v = √(24² + 10²) = √676 = 26 m/s.
Q20 — Projectile Motion · medium · numerical
A projectile is fired at 40 m/s at 60° above the horizontal. Its horizontal range is (g = 10 m/s²):
A. 80 m
B. 160 m
C. 40√3 ≈ 69.3 m
D. 80√3 ≈ 138.6 m ✓ Correct
Solution: R = u²sin2θ/g = (1600 × sin120°)/10 = 160 × (√3/2) = 80√3 ≈ 138.6 m.