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Projectile Motion (NCERT) - Free MCQ Practice Test with solutions, NEET


MCQ Practice Test & Solutions: Test: Projectile Motion (NCERT) (25 Questions)

You can prepare effectively for NEET NCERT Based Tests for NEET with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Projectile Motion (NCERT)". These 25 questions have been designed by the experts with the latest curriculum of NEET 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 25 minutes
  • - Number of Questions: 25

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Test: Projectile Motion (NCERT) - Question 1

Which of the following is true regarding projectile motion?

Detailed Solution: Question 1

Horizontal velocity of a projectile is not affected by gravity.

Test: Projectile Motion (NCERT) - Question 2

A bomb is released by a horizontal flying aeroplane. The trajectory of the bomb is

Detailed Solution: Question 2

It is an example of projectile motion. Therefore, the trajectory of the bomb is parabola.

Test: Projectile Motion (NCERT) - Question 3

 In case of a projectile motion, what is the angle between the velocity and acceleration at the highest point?

Detailed Solution: Question 3

At the highest point, velocity is acting horizontally and acceleration ( = acceleration due to gravity) is acting vertically downwards. Therefore, at the highest point the angle between velocity and acceleration is 90 .

Test: Projectile Motion (NCERT) - Question 4

If a body is projected with an angle θ to the horizontal, then

Detailed Solution: Question 4

In angular projection, the body at the highest point has velocity = u cosθ in the horizontal direction which makes zero angles with horizontal direction.

Test: Projectile Motion (NCERT) - Question 5

Two particles are projected simultaneously in the same vertical plane, from the same point, both with different speeds and at different angles with horizontal. The path followed by one, as seen by the other, is

Detailed Solution: Question 5

Letbe the initial velocities of the two particles and θ1​ and θ2​ be their angles of projection with the horizontal.
The velocities of the two particles after time t are,



Test: Projectile Motion (NCERT) - Question 6

A football is kicked into the air vertically upwards with velocity u. The velocity of the ball at the highest point is

Detailed Solution: Question 6

The velocity of the ball at the highest point is zero.

Test: Projectile Motion (NCERT) - Question 7

 The equations of motion of a projectile are given by x = 36t m and 2y = 96t - 9.8t2 m. The angle of projection is

Detailed Solution: Question 7

Given x = 36t
and 2y = 96t − 9.8t2
or y = 48t − 4.9t2
Let the initial velocity of projectile be u and angle of projection θ. Then, Initial horizontal component of velocity,

Test: Projectile Motion (NCERT) - Question 8

The relation between the time of flight of projectile Tf, and the time to reach the maximum height tm is

Detailed Solution: Question 8

Time to reach maximum height = tm
Time to reach back to ground = tm
Total time of flight, Tf = tm + tm = 2tm

Test: Projectile Motion (NCERT) - Question 9

A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with a speed of 15 m s-1. The time taken by the stone to reach the ground is

Detailed Solution: Question 9

Here, u = 15ms−1

h = 490m

Time taken by the stone to reach the ground is

Test: Projectile Motion (NCERT) - Question 10

The speed with which the stone hits the ground is

Detailed Solution: Question 10

Motion along horizontal direction, ↓+ ve
ux = 15ms−1, ax = 0 
vx = ux + axt = 15 + 0 × 10 = 15ms−1
Motion along vertical direction,
uy = 0,ay = g
vy = uy + ayt = 0 + 9.8 × 10 = 98ms−1
∴ Speed of the stone when it hits the ground is

Test: Projectile Motion (NCERT) - Question 11

Two balls are projected at an angle θ and (90° - θ) to the horizontal with the same speed. The ratio of their maximum vertical heights is

Detailed Solution: Question 11

Test: Projectile Motion (NCERT) - Question 12

Two particles are projected in air with speed u at angles θ1​ and θ2​ (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then which one of the following is correct?
(where T1​ and T2​ are the time of flight.)

Detailed Solution: Question 12

Test: Projectile Motion (NCERT) - Question 13

The ceiling of a hall is 40 m high. For maximum horizontal distance, the angle at which the ball may be thrown with a speed of 56 m s-1 without hitting the ceiling of the hall is

Detailed Solution: Question 13

Here, u = 56 m s-1 Let θ be the angle of projection with the horizontal to have maximum range, with maximum height = 40 m

Test: Projectile Motion (NCERT) - Question 14

The maximum horizontal distance covered by the ball will be

Detailed Solution: Question 14

Test: Projectile Motion (NCERT) - Question 15

If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is:

Detailed Solution: Question 15

Test: Projectile Motion (NCERT) - Question 16

When air resistance is taken into account while dealing with the motion of the projectile which of the following properties of the projectile, shows an increase?

Detailed Solution: Question 16

In the presence of air resistance, the range, maximum height, speed at which the projectile strikes the ground will decreases whereas the angles at which the projectile strikes the ground will increases.

Test: Projectile Motion (NCERT) - Question 17

Two projectiles are are fired from the same point with the same speed at angles 60° and 30° respectively. Which one of the follwing is true?

Detailed Solution: Question 17

The horizontal range is the same when angle of projection is θ or (90∘-θ).

Test: Projectile Motion (NCERT) - Question 18

Galileo writes that for angles of projection of a projectile at angle (45o + θ) and (45o − θ), the horizontal ranges described by the projectile are in the ratio of: (if θ ≤ 45o)

Detailed Solution: Question 18

Test: Projectile Motion (NCERT) - Question 19

 A cricket ball is thrown at a speed of 30ms−1 in a direction 30 above the horizontal. The time taken by the ball to return to the same level is
(Take g = 10ms−2)

Detailed Solution: Question 19

Here, u = 30ms−1,

θ = 30,

g = 10ms−2

The time taken by the ball to return to the same level is

Test: Projectile Motion (NCERT) - Question 20

 A cricket ball is thrown at a speed of 30ms−1 in a direction 30 above the horizontal. Then the distance from the thrower to the point where the ball returns to the same level is

Detailed Solution: Question 20

Given data,

u = 30ms−1

θ = 300

g = 10ms−2

The distance from the thrower to the point where the ball returns to the same level is

Test: Projectile Motion (NCERT) - Question 21

 A cricket ball is thrown at a speed of 30ms−1 in a direction 30∘ above the horizontal. The maximum height attained by the ball is

Detailed Solution: Question 21

The maximum height is given by

Test: Projectile Motion (NCERT) - Question 22

A cricketer can throw a ball to a maximum horizontal distance of 100m. With the same speed how much high above the ground can the cricketer throw the same ball?

Detailed Solution: Question 22

Let u be the velocity of projection of the ball. The ball will cover maximum horizontal distance when angle of projection with horizontal, θ = 45. Then Rmax = u2/g = 100m
If ball is projected vertically upwards (θ = 90) from ground then H attains maximum value.

∴  The height to which cricketer can through the ball is = Rmax/2 = 100/2 = 50m.

Test: Projectile Motion (NCERT) - Question 23

An aeroplane flying horizontally with a speed of 360 km h-1 releases a bomb at a height of 490 m from the ground. If g = 9.8 m s-2, it will strike the ground at

Detailed Solution: Question 23

Time taken by the bomb to fall through a height of 490 m

Distance at which the bomb strikes the ground = horizontal velocity x time

Test: Projectile Motion (NCERT) - Question 24

A ball is thrown from the top of a tower with an initial velocity of 10 m s-1 at an angle ot 30° with the horizontal. If it hits the ground at a distance of 17.3 m from the base of the tower, the height of the tower is (Take g = 10 m s-2)

Detailed Solution: Question 24

The ball is thrown at an angle, θ = 30o.

Initial velocity of the ball, u = 10 m/s 

Horizontal range of the ball, R = 17.3 m
We know that, R = u cosθ t,
 where t is the time of flight 

using equation of motion we get:-

⟹ Height of tower, h = 10 m

Test: Projectile Motion (NCERT) - Question 25

The speed of a projectile at its maximum height is √3/2 times its initial speed. If the range of the projectile is P times the maximum height attained by it, then P equals

Detailed Solution: Question 25

Given √3/2 ​​u = u cos θ
speed at maximum height or cos θ = √3/2 or θ = 30o

Given that P Hmax ​=  R

We know Hmax ​= R tan θ​/4

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