Projectile Motion Important Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. When do we get maximum height in projectile motion?

1. When θ = 90°
2. When θ = 45°
3. When θ = 0°
4. When θ = 60°

Ans. The correct answer is a. When θ = 90°

Explanation: The formula for the maximum height is given by

h_max = (u sinθ)²/2g

This will be maximum when sin θ = 1, which implies that θ = 90°.

Ques. When do we get the maximum range in projectile motion?

1. When θ = 0°
2. When θ = 45°
3. When θ = 90°
4. When θ = 60°

Ans. The correct answer is b. When θ = 45°

Explanation: The horizontal range in projectile motion is given by

R = u²(sin 2θ)/g

This will be maximum when sin 2θ = 1

⇒ 2θ = 90°

⇒ θ = 45°.

Ques. At what angle of projectile (θ) is the horizontal range minimum?

1. θ = 75°
2. θ = 60°
3. θ = 45°
4. θ = 90°

Ans. The correct answer is d. θ = 90°

Explanation: The horizontal range in projectile motion is given by

R = u²(sin 2θ)/g

When is equal to θ = 90°, sin 2θ = sin 180° = 0. As a result, the range covered becomes 0. Because the range cannot be negative, 0 is the minimum value it can achieve.

Ques. In a normal projectile motion, which of the following quantities has an effect in calculating the body’s mass?

1. Time
2. Horizontal range
3. Force
4. Velocity

Ans. The correct answer is c. Force

Explanation: Except for force, all of the mentioned factors are kinematic. Because Force = m x a is a kinetic quantity, the mass of the body affects the force.

Ques. A body of mass m, projected at an angle of θ from the ground with an initial velocity of u, acceleration due to gravity is g, what is the maximum horizontal range covered?

1. R = u² (sin 2θ)/g
2. R = u² (sin 2θ)/2g
3. R = u² (sin θ)/2g
4. R = u² (sin θ)/g

Ans. The correct answer is a. R = u² (sin 2θ)/g

Explanation: The maximum horizontal range is given by the formula

R = u² (sin 2θ)/g

Ques. The path traced by a projectile in space is known as

1. Orbit
2. Coral
3. Trajectory
4. Track

Ans. The correct answer is c. Trajectory

Explanation: The path traced by a projectile in space is known as a trajectory. The trajectory of a projectile is a parabolic curve that is determined by the initial velocity of the projectile, the angle of projection, and the acceleration due to gravity.

Ques. To take the longest possible jump, an athlete should make an angle of

1. 30 degrees with the ground
2. 45 degrees with the ground
3. 60 degrees with the ground
4. 90 degrees with the ground

Ans. The correct answer is b. 30 degrees with the ground

Explanation: To take the longest possible jump, an athlete should make an angle of 45 degrees with the ground.

Ques. Which of the following is not a projectile motion

1. A stone is thrown horizontally from a building
2. A bullet fired from a gun
3. A stone is thrown in any direction
4. A car moving in a straight line

Ans. The correct answer is d. A car moving in a straight line

Explanation: The car moving in a straight line is not a projectile motion.

Ques. A stone has just been thrown out of a train’s window as it travels along a horizontal straight track. Following which path, the stone will fall to the ground.

1. Parabolic path
2. Hyperbolic path
3. Circular path
4. Straight path

Ans. The correct answer is a. Parabolic path

Explanation: When a stone is thrown out of a moving train, it retains the horizontal velocity of the train while also being acted upon by gravity, which pulls it downward. This combination of horizontal and vertical motion causes the stone to follow a parabolic path, much like the path of a projectile launched from the ground.

Ques. When a bullet is fired horizontally, another bullet is dropped from the same height. They will hit the ground

1. at different times depending on the observer
2. at the same time
3. One after the other
4. None of the above

Ans. The correct answer is b. at the same time

Explanation: Both bullets will hit the ground at the same time, regardless of the observer's reference frame. This is because the vertical motion of the bullets is independent of their horizontal motion. In other words, the horizontal velocity of the bullet that is fired horizontally does not affect its vertical velocity. As a result, both bullets will experience the same acceleration due to gravity and will reach the ground at the same time.

Ques. An airplane drops a bomb while moving horizontally at a constant speed. When air resistance is taken into account, the bomb

1. falls to earth behind the plane
2. falls to earth exactly below the plane
3. flies alongside the plane
4. falls to earth ahead of the plane

Ans. The correct answer is a. fall to earth behind the plane

Explanation: When a bomb is dropped from an airplane moving horizontally at a constant speed, it initially has the same horizontal velocity as the airplane. However, as the bomb falls, it is also acted upon by air resistance, which slows it down horizontally. This means that the bomb will not travel as far horizontally as the airplane, and it will fall to the ground behind the airplane.

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Short Answers Questions [2 Marks Questions]

Ques. What is Projectile Motion?

Ans. When a particle is thrown horizontally near the surface of the earth, it moves along a curved path with constant acceleration. This curved path is always directed to the center of the Earth. The path of such a particle is known as the trajectory of the projectile, and the motion is known as projectile motion

Ques. What is the trajectory?

Ans. A trajectory is the path that a projectile follows. When an object is thrown into space and the only force acting on it is gravity, it is referred to as a projectile. This does not imply that other forces aren't acting, but gravity is the primary force that acts on a projectile. Therefore, the impacts of other forces are minimized.

Ques. What is the time of flight?

Ans. The time of flight of a projectile motion is defined as the time taken during the time the object is projected and the time it reaches the surface.

Ques. What is the concept of acceleration in horizontal and vertical projectile motion?

Ans. When an object is thrown in the air at a certain speed, the only force acting on it is acceleration due to gravity (g). Because it acts vertically downwards, there is no acceleration in the horizontal direction, hence the velocity of the particle remains constant in the horizontal direction.

Ques. What is the condition for maximum range in a normal projectile?

Ans. The formula for the horizontal range is 

R = u²(sin 2θ/g) 

Therefore, when sin 2θ = 1, the value of R will be maximum, indicating that 2θ = 90°, implying that should be 45°.

Ques. What is meant by projectile?

Ans. Any object that is cast, fired, flung, heaved, hurled, pitched, tossed, or thrown is considered a projectile.

Throwing a ball straight up, kicking a ball at an angle to the horizontal, or simply dropping something and allowing it to fall are all examples of projectile motion. 

Ques. What assumptions are applied when studying projectile motion?

Ans. The following are the assumptions applied when studying projectile motion

• Because air has no frictional resistance, we assume the external resistance is 0 unless otherwise specified.
• Because the impact of the earth's rotation and curvature is minor, it is ignored.
• Acceleration due to gravity is constant in magnitude and direction at all points of the projectile path.

Ques. What are the real-life examples of projectile motion?

Ans. The following are real-life examples of projectile motion

• When a cannonball launches from a cannon, it follows a curved path rather than a straight line. This is due to the shot's angle, which allows the ball to go vertically and horizontally at the same time.
• During practice, a javelin thrower directs the javelin's sharp edge in the air at a certain angle. The javelin's initial velocity has both horizontal and vertical components. When the vertical velocity approaches zero or the maximum height is attained, the javelin travels horizontally.

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Very Long Answers Questions [3 Marks Questions]

Ques. A projectile is fired horizontally with a velocity of 98 m/s from the top of a hill 490 m high. Find

1. The velocity with which the projectile strikes the ground.
2. Time taken to reach the ground.
3. The distance of the target from the hill.

Ans. Given

• The initial velocity of the projectile, u = 98 m/s
• Height of the hill, h = -490 m

1. The velocity with which the projectile strikes the ground is given by

v = √(u² + 2gh)

⇒ v = √(98² + 2 x 9.8 x 490) = 138.57 m/s

2. Time taken by the projectile to reach the ground is given by

T = √(2h/g)

⇒ T = √(2 x 490 / 9.8) = 10 seconds

100. The distance of the target from the hill is given by

R = uT

⇒ R = 98 x 10 = 980 m

Ques. How much high above the ground a boy can throw the ball if he is able to throw the same ball up to a maximum horizontal distance of 50 m?

Ans. Given, the boy can able to throw the ball to a maximum horizontal distance of 50 m. Therefore 

Maximum horizontal range, R_max = 50 m

Also, R_max = u²/g

Where u is the initial velocity of the ball

⇒ u²/g = 50 …(i)

For the vertical motion of the ball, Use the equation of motion

v² = u² + 2(-g)h

Where 

• v is the velocity at the highest point
• h is the maximum vertical height covered by the ball
• g is the acceleration due to gravity

We have, v = 0 (at maximum height)

⇒ u² = 2gh

⇒ 2h = u²/g

⇒ 2h = 50 [from equation (i)]

⇒ h = 25 m

Ques. A ball is thrown with a speed of 20 m/s at an elevation angle of 45°. Find its time of flight and the horizontal range. [Take g = 10 m/s²]

Ans. Given

• Initial velocity of the ball, u = 20 m/s
• Elevation angle i.e. the angle through which the ball is thrown, θ = 45°

The time of flight of a body in projectile motion is given by

T = (2u sinθ)/g

On substituting the values, we get

T = (2 x 20 x sin 45°)/10 = 2.828 seconds

The formula of the horizontal range is given by

R = (u² sin2θ)/g

On substituting the values, we get

R = (20² x sin 90°)/10 = 40 m

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Long Answers Questions [5 Marks Questions]

Ques. Show that the path followed by a projectile is parabolic.

Ans. Let a body be projected with speed u m/s at an angle θ with the horizontal.

The motion of the projectile is in plane and the equation of this motion is given by

x = v_oxt + 1/2a_xt²

y = v_oyt + 1/2a_yt²

For horizontal motion: 

• v_ox = u
• a_x = 0

On substituting the values, we get

x = ut

⇒ t = x/u …(i)

For vertical motion:

• v_oy = 0
• a_y = -g
• y = -y

On substituting the values, we get

-y = 0 - (1/2)gt²

Using equation (i), 

⇒ y = (1/2)g(x/u)²

⇒ y = (g/2u²)x²

⇒ y = kx²

Where k = g/2u², is constant

The equation y = kx² is the equation of parabola. Thus the path followed by a projectile is parabolic.

Ques. A projectile has a range of 40 m and reaches a maximum height of 10 m. Find the angle at which the projectile is fired.

Ans. Given

• The range of the projectile, R = 40 m
• The maximum height of the projectile, H_max = 10 m

The formula of the horizontal range is given by

R = (u² sin²θ)/g

On substituting the value, we get

(u² sin²θ)/g = 40 ….(i)

The formula of the maximum height is given by

H_max = (u² sin²θ)/2g

On substituting the value, we get

(u² sin²θ)/2g = 10 ….(ii)

Dividing equation (i) from equation (ii), we get

[(u² sin²θ)/g]/[(u² sin²θ)/2g] = 40/10

On solving, we get

sinθ/cosθ = 1 ⇒ tanθ = 1

⇒ θ = 45°

Ques. A stone is thrown with a speed of 10 m/s at an angle of projection 60°. Find its height above the point of projection when it is at a horizontal distance of 3 m from the tower. [Take g = 10 m/s²]

Ans. Given

• The initial velocity of the stone, u = 10 m/s
• The angle of projection, θ = 60°
• Horizontal distance, x = 3 m

Using the equation of motion for horizontal motion

x = u_xt + (1/2)a_xt²

Here

• x = 3 m
• u_x = u cosθ
• a_x = 0

On substituting the values, we get

3 = (u cosθ)t + 0

⇒ 3 = (10 cos60°)t

⇒ t = 3/5 seconds

Now considering the vertical motion at time t = 3/5 seconds

y = u_yt + (1/2)a_yt²

Here

• t = 3/5 s
• u_y = u sinθ
• a_y = -g = – 10 m/s²

On substituting the values, we get

y = (u sinθ)t + (1/2)a_yt²

⇒ y = (10 sin60)(3/5) + (1/2)(-10)(3/5)²

⇒ y = 3.396 m

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