Projectile Motion — JEE Main Physics MCQs with Solutions
Free JEE Main 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 projectile is launched with velocity u = (20 î + 20 ĵ) m/s (g = 10 m/s²). Find its maximum height.
A. 40 m
B. 20 m ✓ Correct
C. 10 m
D. 80 m
Solution: H = u_y²/(2g) = 20²/(2×10) = 400/20 = 20 m.
Q2 — Projectile Motion · easy · numerical
A shell is fired at 20 m/s at 37° above the horizontal (sin 37° = 0.6, g = 10 m/s²). Find its time of flight.
A. 3.2 s
B. 4 s
C. 1.2 s
D. 2.4 s ✓ Correct
Solution: T = 2u sin θ/g = 2×20×0.6/10 = 2.4 s.
Q3 — Projectile Motion · easy · numerical
A ball is projected at 40 m/s at 45° to the horizontal (g = 10 m/s²). Find its horizontal range.
A. 160 m ✓ Correct
B. 320 m
C. 120 m
D. 80 m
Solution: R = u² sin 2θ/g = 1600 × sin 90°/10 = 1600/10 = 160 m.
Q4 — Projectile Motion · easy · numerical
A stone is thrown horizontally at 15 m/s from the top of a 20 m tower (g = 10 m/s²). How far from the base of the tower does it land?
A. 15 m
B. 30 m ✓ Correct
C. 40 m
D. 60 m
Solution: Fall time t = √(2h/g) = √(2×20/10) = 2 s, so horizontal distance = 15×2 = 30 m.
Q5 — Projectile Motion · easy · numerical
A ball is thrown horizontally at 8 m/s from a tall building (g = 10 m/s²). Find its speed 1.5 s after it is thrown.
A. 23 m/s
B. 15 m/s
C. 12.5 m/s
D. 17 m/s ✓ Correct
Solution: After 1.5 s, vₓ = 8 m/s and v_y = gt = 10×1.5 = 15 m/s, so speed = √(64 + 225) = 17 m/s.
Q6 — Projectile Motion · easy · numerical
A projectile is fired at 60 m/s at 30° above the horizontal (g = 10 m/s²). Find its maximum height.
A. 45 m ✓ Correct
B. 30 m
C. 90 m
D. 180 m
Solution: u_y = 60 sin 30° = 30 m/s, so H = u_y²/(2g) = 900/20 = 45 m.
Q7 — Projectile Motion · medium · numerical
The trajectory of a projectile is y = x − x²/20, with x and y in metres. Find its maximum height.
A. 5 m ✓ Correct
B. 20 m
C. 10 m
D. 2.5 m
Solution: y = 0 at x = 0 and x = 20, so the range is 20 m; maximum height occurs at x = 10: y = 10 − 100/20 = 5 m.
Q8 — Projectile Motion · medium · numerical
A projectile follows the path y = √3·x − x²/10 (metres). Find its horizontal range.
A. 5√3 ≈ 8.7 m
B. 10 m
C. 20√3 ≈ 34.6 m
D. 10√3 ≈ 17.3 m ✓ Correct
Solution: Setting y = 0: x(√3 − x/10) = 0, so the range is x = 10√3 ≈ 17.3 m.
Q9 — Projectile Motion · medium · numerical
The equation of a projectile's path is y = x − x²/40 (metres), with g = 10 m/s². Find the speed of projection.
A. 10 m/s
B. 10√2 ≈ 14.1 m/s
C. 20 m/s ✓ Correct
D. 40 m/s
Solution: Comparing with y = x tan θ − gx²/(2u² cos²θ): tan θ = 1 (θ = 45°) and 10/(2u²×½) = 1/40, so u² = 400 and u = 20 m/s.
Q10 — Projectile Motion · medium · numerical
A projectile is launched with u = (16 î + 30 ĵ) m/s (g = 10 m/s²). Find its horizontal range.
A. 48 m
B. 96 m ✓ Correct
C. 45 m
D. 192 m
Solution: T = 2u_y/g = 2×30/10 = 6 s, so R = uₓ × T = 16×6 = 96 m.
Q11 — Projectile Motion · medium · numerical
A stone is projected with u = (10 î + 25 ĵ) m/s (g = 10 m/s²). At what times is it at a height of 30 m?
A. 1 s and 4 s
B. 1.5 s and 3.5 s
C. 2.5 s only
D. 2 s and 3 s ✓ Correct
Solution: 30 = 25t − 5t² gives t² − 5t + 6 = 0, i.e. (t − 2)(t − 3) = 0, so t = 2 s (going up) and t = 3 s (coming down).
Q12 — Projectile Motion · medium · numerical
A projectile is launched with u = (24 î + 27 ĵ) m/s (g = 10 m/s²). Find its speed 2 s after projection.
A. 17 m/s
B. 25 m/s ✓ Correct
C. 36.1 m/s
D. 27 m/s
Solution: At t = 2 s, vₓ = 24 m/s and v_y = 27 − 10×2 = 7 m/s, so speed = √(576 + 49) = √625 = 25 m/s.
Q13 — Projectile Motion · medium · numerical
A projectile launched at 53° above the horizontal attains a maximum height of 20 m (tan 53° = 4/3). Find its horizontal range.
A. 45 m
B. 60 m ✓ Correct
C. 106.7 m
D. 80 m
Solution: R = 4H/tan θ = 4×20/(4/3) = 80 × 3/4 = 60 m.
Q14 — Projectile Motion · medium · numerical
A ball is projected with u = (10 î + 30 ĵ) m/s (g = 10 m/s²). What angle does its velocity make with the horizontal 2 s after launch?
A. 45° ✓ Correct
B. 72°
C. 60°
D. 30°
Solution: At t = 2 s, vₓ = 10 m/s and v_y = 30 − 20 = 10 m/s, so tan φ = 10/10 = 1 and φ = 45°.
Q15 — Projectile Motion · medium · numerical
A particle is projected at 30 m/s at 60° to the horizontal (g = 10 m/s²). After what time is its velocity perpendicular to its initial velocity?
A. 6 s
B. 2√3 ≈ 3.46 s ✓ Correct
C. 3 s
D. √3 ≈ 1.73 s
Solution: Velocity becomes perpendicular to the initial velocity at t = u/(g sin θ) = 30/(10 × √3/2) = 6/√3 = 2√3 ≈ 3.46 s.
Q16 — Projectile Motion · medium · numerical
A stone is thrown from the top of a 15 m cliff at 20 m/s at 30° above the horizontal (g = 10 m/s²). How far from the base of the cliff does it land?
A. 15√3 ≈ 26 m
B. 30√3 ≈ 52 m ✓ Correct
C. 30 m
D. 60 m
Solution: Vertically: −15 = 10t − 5t² gives t² − 2t − 3 = 0, so t = 3 s; horizontally, distance = 20 cos 30° × 3 = 10√3 × 3 = 30√3 ≈ 52 m.
Q17 — Projectile Motion · medium · numerical
Two shells fired at the same speed at two different angles both have a range of 45 m (g = 10 m/s²). Find the product of their times of flight.
A. 4.5 s²
B. 3 s²
C. 9 s² ✓ Correct
D. 90 s²
Solution: For equal ranges at complementary angles, T₁T₂ = 2R/g = 2×45/10 = 9 s².
Q18 — Projectile Motion · medium · numerical
A ball is thrown horizontally at 20 m/s from a height of 60 m (g = 10 m/s²). At what angle below the horizontal does it strike the ground?
A. 30°
B. 60° ✓ Correct
C. 45°
D. 53°
Solution: At the ground v_y = √(2gh) = √(2×10×60) = 20√3 m/s, so tan φ = 20√3/20 = √3 and φ = 60°.
Q19 — Projectile Motion · medium · numerical
A ball rolls off a 20 m high roof horizontally at 21 m/s (g = 10 m/s²). Find its speed just before hitting the ground.
A. 29 m/s ✓ Correct
B. 21 m/s
C. 41 m/s
D. 20 m/s
Solution: v_y = √(2gh) = √(2×10×20) = 20 m/s, so speed = √(21² + 20²) = √(441 + 400) = 29 m/s.
Q20 — Projectile Motion · medium · numerical
Two projectiles are fired at 40 m/s, one at 60° and the other at 30° above the horizontal (g = 10 m/s²). Find the difference in their maximum heights.
A. 40 m ✓ Correct
B. 60 m
C. 20 m
D. 80 m
Solution: H = u² sin²θ/(2g): at 60°, H = 1600×0.75/20 = 60 m; at 30°, H = 1600×0.25/20 = 20 m; difference = 40 m.