Projectile Motion — Basics — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Projectile Motion — Basics MCQs with step-by-step solutions (21 questions). Part of Motion in a Plane (Std 11). Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Projectile Motion — Basics · easy · theory
The path followed by a projectile in the absence of air resistance is:
A. A straight line
B. A circle
C. A hyperbola
D. A parabola ✓ Correct
Solution: Uniform horizontal motion combined with uniformly accelerated vertical motion gives a parabola.
Q2 — Projectile Motion — Basics · easy · theory
During projectile motion, the horizontal component of velocity:
A. Becomes zero at the top
B. Remains constant throughout ✓ Correct
C. Increases steadily
D. Decreases steadily
Solution: There is no horizontal force, so there is no horizontal acceleration.
Q3 — Projectile Motion — Basics · easy · theory
During projectile motion, the vertical component of velocity:
A. Remains constant
B. Increases steadily throughout
C. Changes uniformly under gravity ✓ Correct
D. Is always zero
Solution: It decreases on the way up, is zero at the top and increases on the way down.
Q4 — Projectile Motion — Basics · medium · theory
At the highest point of its trajectory, a projectile has:
A. Zero horizontal velocity
B. Zero velocity altogether
C. Zero vertical velocity but non-zero horizontal velocity ✓ Correct
D. Maximum vertical velocity
Solution: Only the vertical component vanishes; the body continues moving horizontally.
Q5 — Projectile Motion — Basics · medium · theory
The horizontal and vertical motions of a projectile are:
A. Both uniform
B. Independent of each other ✓ Correct
C. Coupled through the acceleration
D. Both uniformly accelerated
Solution: This independence is what allows the two components to be treated separately.
Q6 — Projectile Motion — Basics · medium · theory
Throughout the flight of a projectile, its acceleration is:
A. Constant, equal to $g$ and directed vertically downward ✓ Correct
B. Variable in magnitude
C. Directed along the velocity
D. Zero at the highest point
Solution: Gravity acts uniformly throughout, including at the topmost point.
Q7 — Projectile Motion — Basics · medium · theory
The time of flight of a projectile launched with speed $u$ at angle $\theta$ on level ground is:
A. $\dfrac{u^2\sin 2\theta}{g}$
B. $\dfrac{u\sin\theta}{g}$
C. $\dfrac{2u\cos\theta}{g}$
D. $\dfrac{2u\sin\theta}{g}$ ✓ Correct
Solution: It is twice the time taken to reach the highest point.
Q8 — Projectile Motion — Basics · easy · theory
For a body projected horizontally from a height, the initial vertical velocity is:
A. Zero ✓ Correct
B. Maximum
C. Equal to $g$
D. Equal to the horizontal velocity
Solution: The body falls exactly as though dropped from rest, while moving horizontally at constant speed.
Q9 — Projectile Motion — Basics · medium · numerical
A projectile is launched at $20\text{ m/s}$ at $30^\circ$ to the horizontal. Its time of flight is ($g = 10\text{ m/s}^2$):
A. $4\text{ s}$
B. $3.46\text{ s}$
C. $1\text{ s}$
D. $2\text{ s}$ ✓ Correct
Solution: $T = \dfrac{2u\sin\theta}{g} = \dfrac{2 \times 20 \times 0.5}{10} = 2\text{ s}$.
Q10 — Projectile Motion — Basics · hard · numerical
A ball is thrown horizontally at $20\text{ m/s}$ from a tower $80\text{ m}$ high. It strikes the ground at a horizontal distance of ($g = 10\text{ m/s}^2$):
A. $160\text{ m}$
B. $40\text{ m}$
C. $100\text{ m}$
D. $80\text{ m}$ ✓ Correct
Solution: Time of fall $= \sqrt{\dfrac{2 \times 80}{10}} = 4\text{ s}$, so the range is $20 \times 4 = 80\text{ m}$.
Q11 — Projectile Motion — Basics · medium · numerical
A body is projected horizontally from a height of $80\text{ m}$. Its time of flight is ($g = 10\text{ m/s}^2$):
A. $2\text{ s}$
B. $8\text{ s}$
C. $4\text{ s}$ ✓ Correct
D. $16\text{ s}$
Solution: The vertical motion is free fall: $80 = \dfrac{1}{2}(10)t^2$ gives $t = 4\text{ s}$.
Q12 — Projectile Motion — Basics · medium · numerical
At the highest point of its path, the speed of a projectile launched at $u$ and angle $\theta$ is:
A. $u$
B. Zero
C. $u\cos\theta$ ✓ Correct
D. $u\sin\theta$
Solution: Only the horizontal component survives at the top.
Q13 — Projectile Motion — Basics · medium · numerical
A projectile is launched at $20\text{ m/s}$ at $30^\circ$. Its horizontal component of velocity is approximately:
A. $20\text{ m/s}$
B. $10\text{ m/s}$
C. $17.3\text{ m/s}$ ✓ Correct
D. $14.1\text{ m/s}$
Solution: $u\cos 30^\circ = 20 \times 0.866 \approx 17.3\text{ m/s}$.
Q14 — Projectile Motion — Basics · hard · numerical
The velocity of a projectile at its maximum height is $\dfrac{\sqrt{3}}{2}$ times its initial speed. The angle of projection is:
A. $75^\circ$
B. $30^\circ$ ✓ Correct
C. $45^\circ$
D. $60^\circ$
Solution: At the top the speed is $u\cos\theta$, so $\cos\theta = \dfrac{\sqrt{3}}{2}$ and $\theta = 30^\circ$.
Q15 — Projectile Motion — Basics · hard · numerical
A projectile is launched at $40\text{ m/s}$ at $45^\circ$. Its time of flight is approximately ($g = 10\text{ m/s}^2$):
A. $8\text{ s}$
B. $2.83\text{ s}$
C. $4\text{ s}$
D. $5.66\text{ s}$ ✓ Correct
Solution: $T = \dfrac{2 \times 40 \times 0.707}{10} \approx 5.66\text{ s}$.
Q16 — Projectile Motion — Basics · hard · numerical
A ball thrown horizontally at $20\text{ m/s}$ from a height of $45\text{ m}$ lands at a horizontal distance of ($g = 10\text{ m/s}^2$):
A. $90\text{ m}$
B. $45\text{ m}$
C. $30\text{ m}$
D. $60\text{ m}$ ✓ Correct
Solution: Time of fall $= \sqrt{\dfrac{90}{10}} = 3\text{ s}$, so the range is $20 \times 3 = 60\text{ m}$.
Q17 — Projectile Motion — Basics · medium · numerical
The acceleration of a projectile at the highest point of its trajectory is:
A. $g$, directed vertically downward ✓ Correct
B. $g$, directed horizontally
C. Zero
D. $g$, directed vertically upward
Solution: Gravity never switches off, even where the vertical velocity is momentarily zero.
Q18 — Projectile Motion — Basics · medium · numerical
A body is thrown vertically upward at $10\text{ m/s}$. Its total time of flight is ($g = 10\text{ m/s}^2$):
A. $0.5\text{ s}$
B. $4\text{ s}$
C. $1\text{ s}$
D. $2\text{ s}$ ✓ Correct
Solution: This is the special case $\theta = 90^\circ$: $T = \dfrac{2u}{g} = 2\text{ s}$.
Q19 — Projectile Motion — Basics · easy · numerical
A projectile launched at angle $\theta$ with speed $u$ has a vertical component of initial velocity equal to:
A. $u\sin\theta$ ✓ Correct
B. $u$
C. $u\cos\theta$
D. $u\tan\theta$
Solution: The vertical component governs the height reached and the time of flight.
Q20 — Projectile Motion — Basics · hard · numerical
Two balls are projected horizontally from the same height with different speeds. They reach the ground:
A. The slower one first
B. The faster one first
C. At the same time, since the vertical motions are identical ✓ Correct
D. At times proportional to their speeds
Solution: The horizontal velocity has no influence on the vertical free fall.
Q21 — Projectile Motion — Basics · hard · numerical
A stone is projected at $30^\circ$ with speed $u$. The ratio of its horizontal to vertical component of initial velocity is:
A. $1 : \sqrt{3}$
B. $\sqrt{3} : 1$ ✓ Correct
C. $1 : 1$
D. $2 : 1$
Solution: $\dfrac{u\cos 30^\circ}{u\sin 30^\circ} = \dfrac{0.866}{0.5} = \sqrt{3}$.