Projectile Motion — Range & Height — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Projectile Motion — Range & Height 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 — Range & Height · easy · theory
The horizontal range of a projectile launched with speed $u$ at angle $\theta$ is:
A. $\dfrac{u^2\sin\theta}{g}$
B. $\dfrac{u^2\sin 2\theta}{g}$ ✓ Correct
C. $\dfrac{2u\sin\theta}{g}$
D. $\dfrac{u^2\sin^2\theta}{g}$
Solution: It is the product of the horizontal velocity and the time of flight.
Q2 — Projectile Motion — Range & Height · easy · theory
The horizontal range of a projectile is maximum when the angle of projection is:
A. $30^\circ$
B. $45^\circ$ ✓ Correct
C. $90^\circ$
D. $60^\circ$
Solution: $\sin 2\theta$ reaches its maximum of one when $2\theta = 90^\circ$.
Q3 — Projectile Motion — Range & Height · medium · theory
The maximum height reached by a projectile is:
A. $\dfrac{u^2\sin 2\theta}{2g}$
B. $\dfrac{u^2\sin\theta}{2g}$
C. $\dfrac{u\sin^2\theta}{2g}$
D. $\dfrac{u^2\sin^2\theta}{2g}$ ✓ Correct
Solution: It follows from $v^2 = u^2 - 2gH$ applied to the vertical component.
Q4 — Projectile Motion — Range & Height · medium · theory
Two projectiles launched at complementary angles with the same speed have:
A. Nothing in common
B. The same time of flight
C. The same maximum height
D. The same horizontal range ✓ Correct
Solution: $\sin 2\theta = \sin(180^\circ - 2\theta)$, so $30^\circ$ and $60^\circ$ give equal ranges.
Q5 — Projectile Motion — Range & Height · medium · theory
The maximum possible range of a projectile of launch speed $u$ is:
A. $\dfrac{u}{g}$
B. $\dfrac{u^2}{g}$ ✓ Correct
C. $\dfrac{u^2}{2g}$
D. $\dfrac{2u^2}{g}$
Solution: Setting $\theta = 45^\circ$ makes $\sin 2\theta = 1$.
Q6 — Projectile Motion — Range & Height · medium · theory
The maximum height of a projectile is greatest when the angle of projection is:
A. $90^\circ$ ✓ Correct
B. $60^\circ$
C. $45^\circ$
D. $30^\circ$
Solution: A vertical launch puts the entire speed into the vertical component.
Q7 — Projectile Motion — Range & Height · hard · theory
The equation of the trajectory of a projectile is of the form:
A. $y = x\tan\theta + \dfrac{gx^2}{2u^2}$
B. $y = \dfrac{gx^2}{2u^2}$
C. $y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$ ✓ Correct
D. $y = x\sin\theta$
Solution: It is quadratic in $x$, which confirms that the path is a parabola.
Q8 — Projectile Motion — Range & Height · easy · theory
The range of a projectile is independent of:
A. The angle of projection
B. The speed of projection
C. The acceleration due to gravity
D. The mass of the projectile ✓ Correct
Solution: Mass appears nowhere in $R = \dfrac{u^2\sin 2\theta}{g}$.
Q9 — Projectile Motion — Range & Height · hard · theory
For a projectile, the relation between range and maximum height is:
A. $R = 4H\cot\theta$ ✓ Correct
B. $R = 4H\tan\theta$
C. $R = H\cot\theta$
D. $R = 2H\cot\theta$
Solution: Dividing $R = \dfrac{u^2\sin 2\theta}{g}$ by $H = \dfrac{u^2\sin^2\theta}{2g}$ gives this result.
Q10 — Projectile Motion — Range & Height · medium · numerical
A projectile launched at $20\text{ m/s}$ at $45^\circ$ has a horizontal range of ($g = 10\text{ m/s}^2$):
A. $34.6\text{ m}$
B. $20\text{ m}$
C. $40\text{ m}$ ✓ Correct
D. $80\text{ m}$
Solution: $R = \dfrac{u^2\sin 90^\circ}{g} = \dfrac{400}{10} = 40\text{ m}$.
Q11 — Projectile Motion — Range & Height · hard · numerical
A projectile launched at $20\text{ m/s}$ at $30^\circ$ has a horizontal range of approximately ($g = 10\text{ m/s}^2$):
A. $34.6\text{ m}$ ✓ Correct
B. $20\text{ m}$
C. $40\text{ m}$
D. $17.3\text{ m}$
Solution: $R = \dfrac{400 \times \sin 60^\circ}{10} = 40 \times 0.866 \approx 34.6\text{ m}$.
Q12 — Projectile Motion — Range & Height · hard · numerical
A projectile launched at $20\text{ m/s}$ at $30^\circ$ reaches a maximum height of ($g = 10\text{ m/s}^2$):
A. $20\text{ m}$
B. $2.5\text{ m}$
C. $10\text{ m}$
D. $5\text{ m}$ ✓ Correct
Solution: $H = \dfrac{u^2\sin^2\theta}{2g} = \dfrac{400 \times 0.25}{20} = 5\text{ m}$.
Q13 — Projectile Motion — Range & Height · medium · numerical
A body projected vertically upward at $20\text{ m/s}$ reaches a maximum height of ($g = 10\text{ m/s}^2$):
A. $5\text{ m}$
B. $20\text{ m}$ ✓ Correct
C. $10\text{ m}$
D. $40\text{ m}$
Solution: With $\theta = 90^\circ$, $H = \dfrac{u^2}{2g} = \dfrac{400}{20} = 20\text{ m}$.
Q14 — Projectile Motion — Range & Height · medium · numerical
Two projectiles are launched with the same speed at $30^\circ$ and $60^\circ$. Their horizontal ranges are:
A. In the ratio $1 : 3$
B. In the ratio $2 : 1$
C. In the ratio $1 : 2$
D. Equal ✓ Correct
Solution: The angles are complementary, so $\sin 60^\circ = \sin 120^\circ$ makes the ranges identical.
Q15 — Projectile Motion — Range & Height · medium · numerical
The maximum range of a projectile launched at $30\text{ m/s}$ is ($g = 10\text{ m/s}^2$):
A. $90\text{ m}$ ✓ Correct
B. $180\text{ m}$
C. $60\text{ m}$
D. $45\text{ m}$
Solution: $R_{max} = \dfrac{u^2}{g} = \dfrac{900}{10} = 90\text{ m}$.
Q16 — Projectile Motion — Range & Height · easy · numerical
If the launch speed of a projectile is doubled at the same angle, its range becomes:
A. Half as large
B. Four times as large ✓ Correct
C. Unchanged
D. Twice as large
Solution: $R \propto u^2$.
Q17 — Projectile Motion — Range & Height · hard · numerical
A projectile launched at $40\text{ m/s}$ at $30^\circ$ reaches a maximum height of ($g = 10\text{ m/s}^2$):
A. $80\text{ m}$
B. $10\text{ m}$
C. $20\text{ m}$ ✓ Correct
D. $40\text{ m}$
Solution: $H = \dfrac{1600 \times 0.25}{20} = 20\text{ m}$.
Q18 — Projectile Motion — Range & Height · hard · numerical
The horizontal range of a projectile equals its maximum height. The angle of projection is:
A. $\tan^{-1}(2)$
B. $\tan^{-1}(0.25)$
C. $\tan^{-1}(1)$
D. $\tan^{-1}(4)$ ✓ Correct
Solution: Setting $R = H$ in $R = 4H\cot\theta$ gives $\cot\theta = \dfrac{1}{4}$, so $\tan\theta = 4$.
Q19 — Projectile Motion — Range & Height · medium · numerical
A projectile launched at $10\text{ m/s}$ at $45^\circ$ has a range of ($g = 10\text{ m/s}^2$):
A. $20\text{ m}$
B. $5\text{ m}$
C. $10\text{ m}$ ✓ Correct
D. $14.1\text{ m}$
Solution: $R = \dfrac{100 \times 1}{10} = 10\text{ m}$.
Q20 — Projectile Motion — Range & Height · hard · numerical
For a projectile launched at $45^\circ$, the ratio of its range to its maximum height is:
A. $4 : 1$ ✓ Correct
B. $2 : 1$
C. $1 : 1$
D. $1 : 4$
Solution: $R = 4H\cot 45^\circ = 4H$.
Q21 — Projectile Motion — Range & Height · medium · numerical
If the acceleration due to gravity were halved, the range of a projectile launched with the same speed and angle would:
A. Halve
B. Become four times
C. Remain unchanged
D. Double ✓ Correct
Solution: $R \propto \dfrac{1}{g}$.