Projectile Motion Questions and Answers: Important for Class 11 Physics

Projectile motion questions in Class 11 are subjective and reasoning-based: you have to prove a result, explain a situation or solve a short numerical. This page gives 21 such projectile motion questions with answers, in the order of the topic, so you can revise the projectile motion questions chapter-wise. If you need the theory first, read eSaral’s Motion in a Plane notes with projectile motion, and revise vector resolution and motion in two dimensions, because every answer below starts by splitting the velocity into horizontal and vertical components.

Projectile Motion Formulas Used in These Questions

Use these formulas for all projectile motion questions. Take a projectile fired with speed $u$ at an angle $\theta$ with the horizontal, from and to the same level, with no air resistance and a constant $g$.

Projectile Motion Questions on Speed, Velocity and Momentum

These projectile motion questions test how the velocity and the momentum change during the flight.

Q1. At which point of the trajectory is the speed minimum?

At the highest point. The vertical velocity is zero there, so the speed is only the horizontal component, $u\cos\theta$, which is the minimum. For a body projected horizontally from a height, the speed only increases after launch, so the minimum is at the start.

Q2. Name a quantity that remains unchanged during the flight of an oblique projectile.

The horizontal component of velocity, $u\cos\theta$, because no force acts in the horizontal direction. The acceleration ($g$, downward) and the total mechanical energy are also constant when air resistance is ignored.

Q3. Name five physical quantities that change during the motion of an oblique projectile.

The speed, the vertical component of velocity, the direction of velocity, the kinetic energy and the potential energy change continuously. The momentum and the height above the ground also change.

Q4. A projectile of mass $m$ is fired with velocity $u$ at an angle $\theta$ with the horizontal. What is the change in momentum as it rises to the highest point?

The vertical momentum changes from $mu\sin\theta$ to zero, and the horizontal momentum $mu\cos\theta$ does not change.

$$\Delta p = 0 - mu\sin\theta = -mu\sin\theta$$

So the change in momentum has magnitude $mu\sin\theta$ and is directed vertically downward.

Q5. A projectile of mass $m$ is projected with velocity $u$ at an angle $\theta$. What is the magnitude of the change in momentum after time $t$?

The only force is gravity, $mg$ downward. The change in momentum equals the impulse of this force.

$$\Delta p = F \times t = mgt$$

The direction is vertically downward, and the result is valid until the projectile lands.

Q6. What is the change in momentum between the initial and final points of the path, if the range is maximum?

The range is maximum at $\theta = 45^\circ$. The horizontal momentum is the same at the start and the end. The vertical momentum changes from $mu\sin 45^\circ$ upward to $mu\sin 45^\circ$ downward.

$$\Delta p = -2mu\sin 45^\circ = -\sqrt{2}\,mu$$

So the magnitude is $\sqrt{2}\,mu$, directed vertically downward.

Projectile Motion Questions on the Path of a Projectile

The two proofs below are the projectile motion questions that appear most often in school exams.

Q7. A body is projected horizontally with a uniform velocity from the top of a tower. Show that the path is a parabola.

Take the origin at the top of the tower, with $x$ horizontal and $y$ downward. The horizontal motion has no acceleration, and the vertical motion starts from rest with acceleration $g$.

$$x = ut, \qquad y = \tfrac{1}{2}gt^2$$

Eliminate $t = x/u$:

$$y = \frac{g}{2u^2}\,x^2$$

This is of the form $y \propto x^2$, so the path is a parabola.

Q8. Prove that the path of a projectile is a parabola.

For a projectile fired with speed $u$ at an angle $\theta$:

$$x = (u\cos\theta)\,t, \qquad y = (u\sin\theta)\,t - \tfrac{1}{2}gt^2$$

From the first equation, $t = \dfrac{x}{u\cos\theta}$. Substituting in the second,

$$y = x\tan\theta - \frac{g}{2u^2\cos^2\theta}\,x^2$$

This is of the form $y = ax - bx^2$ with constants $a$ and $b$, which is the equation of a parabola.

Q9. What is the path followed by a javelin projected horizontally by an athlete?

A parabola. A body projected horizontally follows half of a parabola, because it has a constant horizontal velocity and a vertical motion under gravity (when air resistance is ignored).

Projectile Motion Questions on Range and Angle of Projection

Projectile motion questions on range and angle use the formula $R = \dfrac{u^2\sin 2\theta}{g}$ again and again.

Q10. What are the possible angles of projection with the same velocity to have the same range?

Two angles that add up to $90^\circ$, that is, $\theta$ and $(90^\circ - \theta)$. Since $R = \dfrac{u^2\sin 2\theta}{g}$ and $\sin 2(90^\circ - \theta) = \sin(180^\circ - 2\theta) = \sin 2\theta$, the range is the same. For example, $30^\circ$ and $60^\circ$, or $15^\circ$ and $75^\circ$.

Q11. Galileo stated that “for elevations which exceed or fall short of $45^\circ$ by equal amounts, the ranges are equal.” Prove this statement.

Let the two elevations be $45^\circ + \alpha$ and $45^\circ - \alpha$.

$$R_1 = \frac{u^2\sin(90^\circ + 2\alpha)}{g} = \frac{u^2\cos 2\alpha}{g}, \qquad R_2 = \frac{u^2\sin(90^\circ - 2\alpha)}{g} = \frac{u^2\cos 2\alpha}{g}$$

So $R_1 = R_2$.

Q12. What is the angle of projection at which the maximum height and the horizontal range are equal?

Put $R = H$:

$$\frac{u^2\sin 2\theta}{g} = \frac{u^2\sin^2\theta}{2g} \;\Rightarrow\; 2\sin\theta\cos\theta = \frac{\sin^2\theta}{2} \;\Rightarrow\; \tan\theta = 4$$

So $\theta = \tan^{-1}4 \approx 76^\circ$.

Q13. If the time of flight of a projectile projected with velocity $u$ at an angle $\theta$ is $T$, find the condition for maximum range and its value.

The time of flight is $T = \dfrac{2u\sin\theta}{g}$, and the range is

$$R = (u\cos\theta)\,T = \frac{u^2\sin 2\theta}{g}$$

The range is maximum when $\sin 2\theta = 1$, that is, $\theta = 45^\circ$, and $R_{max} = \dfrac{u^2}{g}$. At this angle $T = \dfrac{\sqrt{2}\,u}{g}$, so in terms of the time of flight, $R_{max} = \dfrac{gT^2}{2}$.

Q14. Prove that a gun will shoot three times as high when its angle of elevation is $60^\circ$ as when it is $30^\circ$, but cover the same horizontal range.

The height is $H = \dfrac{u^2\sin^2\theta}{2g}$, so

$$\frac{H_{60^\circ}}{H_{30^\circ}} = \frac{\sin^2 60^\circ}{\sin^2 30^\circ} = \frac{3/4}{1/4} = 3$$

The range is $R = \dfrac{u^2\sin 2\theta}{g}$. For $30^\circ$ it is proportional to $\sin 60^\circ$, and for $60^\circ$ it is proportional to $\sin 120^\circ = \sin 60^\circ$. So the ranges are equal.

Projectile Motion Questions on Real-Life Situations

These projectile motion questions ask you to explain a situation in words, so state the reason clearly.

Q15. A hunter aims his gun at a monkey sitting on a tree and fires. Will the bullet hit the monkey?

It depends on what the monkey does. Without gravity, the bullet would move along the line of aim. With gravity, the bullet falls below that line by $\tfrac{1}{2}gt^2$ after time $t$.

• If the monkey stays on the tree, the bullet passes below it, so it misses.
• If the monkey lets go at the instant of firing, it also falls by $\tfrac{1}{2}gt^2$ in the same time, so the bullet hits it, provided the bullet reaches the tree before the monkey lands.

Q16. A bomb is released from a horizontally flying bomber when it is vertically above the target. Shall it hit the target?

No. At the instant of release, the bomb has the same horizontal velocity $v$ as the bomber. It keeps this velocity while it falls, so it follows a parabola and lands ahead of the target, at a horizontal distance $v\sqrt{2h/g}$ from the point below the release. To hit the target, the bomb must be released earlier, when the target is that far ahead.

Q17. Why does a tennis ball bounce higher on hills than on plains?

On hills, the acceleration due to gravity is slightly smaller, and the air is thinner. The maximum height $H = \dfrac{u^2\sin^2\theta}{2g}$ increases when $g$ decreases, and thinner air gives less drag.

Q18. Give two applications and two examples of projectile motion.

Applications: planning the angle and speed of a shell fired from artillery, and sports such as javelin throw, shot put and long jump.

Examples: a football kicked at an angle, and a stone thrown from a building.

Q19. When will the relative velocity of two moving objects be zero?

When both objects move with the same velocity, that is, the same speed in the same direction. For example, two projectiles launched together with the same velocity have the same acceleration $g$ at every instant, so their relative velocity stays zero during the whole flight.

Projectile Motion Numerical Questions With Solutions

Numerical projectile motion questions need the same steps every time: resolve the velocity, find the time, and then find the range or the height.

Q20. The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no atmosphere) are $y = (8t - 5t^2)$ m and $x = 6t$ m, where $t$ is in seconds. Find the velocity and the angle of projection.

At $t = 0$: $v_x = 6\ \text{m/s}$ and $v_y = 8\ \text{m/s}$.

$$v = \sqrt{6^2 + 8^2} = 10\ \text{m/s}, \qquad \theta = \tan^{-1}\frac{8}{6} \approx 53^\circ$$

The acceleration is $a_y = -10\ \text{m/s}^2$, so $g = 10\ \text{m/s}^2$ on this planet. The maximum height is $\dfrac{8^2}{2\times 10} = 3.2$ m, the time of flight is $\dfrac{2\times 8}{10} = 1.6$ s, and the range is $6 \times 1.6 = 9.6$ m.

Q21. A mass is projected horizontally with velocity $v$ from a tower of height $h$. Find the horizontal distance it covers from the foot of the tower.

The time to fall is $t = \sqrt{\dfrac{2h}{g}}$. The horizontal distance is

$$x = vt = v\sqrt{\frac{2h}{g}}$$

Extra numerical 1: Time of flight, height and range

A ball is projected at $30^\circ$ with a speed of $20\ \text{m/s}$. Find $T$, $H$ and $R$. Take $g = 10\ \text{m/s}^2$.

$$T = \frac{2\times 20\times 0.5}{10} = 2\ \text{s}, \qquad H = \frac{20^2\times 0.25}{2\times 10} = 5\ \text{m}, \qquad R = \frac{20^2\sin 60^\circ}{10} \approx 34.6\ \text{m}$$

Extra numerical 2: Projection from a tower

A stone is thrown horizontally at $15\ \text{m/s}$ from the top of an $80\ \text{m}$ tower. Find the time to fall, the horizontal distance and the speed on landing. Take $g = 10\ \text{m/s}^2$.

$$t = \sqrt{\frac{2\times 80}{10}} = 4\ \text{s}, \qquad x = 15\times 4 = 60\ \text{m}$$ $$v_y = gt = 40\ \text{m/s}, \qquad v = \sqrt{15^2 + 40^2} \approx 42.7\ \text{m/s}$$

Practice Projectile Motion Questions With Answers

1. A projectile is fired at $50\ \text{m/s}$ at an angle of $37^\circ$ ($\sin 37^\circ = 0.6$, $\cos 37^\circ = 0.8$). Find $T$, $H$ and $R$. Take $g = 10\ \text{m/s}^2$.
2. The maximum range of a gun is $500\ \text{m}$. Find the speed of the shell. Take $g = 10\ \text{m/s}^2$.
3. A projectile fired at $15^\circ$ has a range $R$. At what other angle will the same speed give the same range?

Answers: 1. $T = 6$ s, $H = 45$ m, $R = 240$ m. 2. $u = \sqrt{gR_{max}} = \sqrt{10\times 500} \approx 71\ \text{m/s}$. 3. $75^\circ$.

For exam-style projectile motion questions, solve the Projectile Motion NEET previous year questions and the JEE Main chapter-wise previous year questions.

Common Mistakes in Projectile Motion Questions

These mistakes cost marks in projectile motion questions.

Related Topics

• Class 11 Physics notes
• Motion in a Plane notes with projectile motion
• Vectors and two-dimensional motion

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