Motion in Plane Numerical on Projectile Motion Class 11 Nootan ISC Physics Solution

Motion in Plane Numerical on Projectile Motion Class 11 Nootan ISC Physics Solution Ch-5. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Motion in Plane Numerical on Projectile Motion Class 11 Nootan ISC Physics Solution

Motion in Plane Numerical on Projectile Motion ISC Class 11 Ch-5 Nootan Physics Solution,  Kumar and Mittal of Nageen Prakashan

Board ISC
Class 11
Subject Physics
Writer Kumar and Mittal
Publication  Nageen Prakashan
Chapter-5 Motion in Plane
Topics Numerical on Projectile Motion
Academic Session 2024-2025

Numerical on Projectile Motion

Motion in Plane Class 11 Nootan ISC Physics Solution Ch-5 Kumar and Mittal Nageen Prakashan

Question-22: Two bullets are fired with horizontal velocities of 50 m/s and 100 m/s from two guns at a height of 19.6 m. (a) Will both the bullets strike the ground? (b) If yes; then after how much time and which bullet will strike first? (c) What would be the path of the bullets? (g = 9.8 m/s²)

Answer- (a) Yes, both the bullets will strike the ground at the same time because the downward acceleration and initial velocity in the downward direction of the two bullets are same. They will take the same time to hit the ground and for half projectile

(b) Height = 19.6m   v=0

h=v t + 1/2 gt^2

=> 19.6 = 0 x t + 1/2 x t^2

=>t^2 = 19.6 / 4.9 = 4,  =>t = 2s

Hence, both bullets will hit the ground simultaneously after 2 s at the same time.

(c) parabola

Question-23: A bomb is fired horizontally with a velocity of 20 m/s from the top of a tower 40 m high. After how much time and at what horizontal distance from the tower will the bomb strike the ground? (g = 9.8 m/s²)

Answer- time taken by bomb to strike ground t =

A bomb is fired horizontally with a velocity of 20 m/s from the top of a tower 40 m high

= 20/7 = 2.86 s

distance covered = speed  x time

= 20 x 2.86 = 57.2 m.

Question-24: A ball is thrown with a velocity of 15 m/s at an angle of 30° with the horizontal. Determine (i) the time of flight of the ball, (ii) the maximum height attained by the ball. (g = 10 m/s²) .

Answer- u = 15 m/s       θ = 30°

(i) T = 2 u sinθ / g = 2 x 15 x sin 30 / 10 = 1.5 s

(ii) H = u² sin² θ / 2g = (15 x 15 x 1/4) / (2 x 10) = 225 /80 = 2.8 m

Question-25: A stone is thrown from the horizontal ground at an angle of projection of 30° with a speed of 49 m/s. Calculate its (i) time of flight, (ii) horizontal range.

Answer- (i) T = 2 u sinθ / g = 2 x 49 x sin 30 / 9.8 = 5 s

(ii) R = u² sin2 θ / g = (49 x 49 x sin 60) / (9.8) = 212.2 m

Question-26: An arrow is thrown in the air. Its time of flight is 5 s and the range is 200 m. Determine (i) the vertical component of the velocity of projection,  (ii) the horizontal component, (iii) maximum height and (iv) the angle made with the horizontal. (g = 9.8 m/s²)

Answer-

An arrow is thrown in the air. Its time of flight is 5 s and the range is 200 m

Question-27: A cannon is to fire up to 500 m horizontally. What should be the angle of projection, if the shells are fired with a velocity of 100 m/s? (g = 10 m/s²)

Answer- R = u² sin2θ / g

500 = (100 x 100 x sin 2θ) / (10)

sin 2θ = 5000/10000

sin 2θ = 1/2 = sin 30°

2θ = 30°

θ = 15° or (90° – 15°) = 75°

supplementary angle have same range.

Question-28: A stone is thrown from a bridge at an angle of 30° down with the horizontal with a velocity of 25 m/s. If the stone strikes the water after 2.5 seconds, then calculate the height of the bridge from the water surface. (g = 9.8 m/s²)

Answer-

A stone is thrown from a bridge at an angle of 30° down with the horizontal with a velocity of 25 m/s

s = u t + 1/2 a t²

h = u t + 1/2 g t²

h = 25 sin 30 x 2.5 + 1/2 x 9.8 (2.5)²

= 12.5 x 2.5 + 4.9 x 6.25

= 61.9 m

Question-29: A stone is thrown from the top of a tower at an angle of 30° up with the horizontal with a velocity of 16 m/s. After 4 seconds of flight it strikes the ground. Calculate the height of the tower from the ground and the horizontal range of the stone. (g = 9.8 m/s²)

Answer-

A stone is thrown from the top of a tower at an angle of 30° up with the horizontal with a velocity of 16 m/s

for vertical position

u = 16 x sin 30 = 8 m/s

h =?   t = 4s

h = -8 x 4 + 1/2 x 9.8 x 4 x 4 = 46.4 m

horizontal range

=  horizontal speed x time

= 16 x cos 30 x 4  = 55.4 m

Question-30: A player throws a ball at an angle of 30° up with the horizontal with a velocity of 14 m/s. If the point of projection is at a height of 12 m from the ground, then calculate the distance up to which the ball is thrown by the player. (g = 10 m/s²)

Answer-

A player throws a ball at an angle of 30° up with the horizontal with a velocity of 14 m/s. If the point of projection is at a height of 12 m from the ground, then calculate the distance up to which the ball is thrown by the player

Question-31: A projectile has a range of 50 m and reaches a maximum height of 10 m. Calculate the angle at which the projectile is fired.

Answer-

A projectile has a range of 50 m and reaches a maximum height of 10 m. Calculate the angle at which the projectile is fired.

Question-32: Find the angle of projection for which the horizontal range and the maximum height are equal.

Answer- R = u² sin2θ / g   ,    H = u² sin² θ / 2g

u² sin2θ / g = u² sin² θ / 2g

2 sin θ cos θ = sin²θ / 2

u cos θ = sin θ

tan θ = 4

θ = tan^-1 (4)

Question-33: A projectile is fired horizontally with a velocity of 98 m/s from the top of a hill 490 m high. Find (i) the time taken to reach the ground (ii) the distance of the target from the hill.

Answer-

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

(i) The projectile is fired from the top 0 of a hill with speed ux = 98 m/s and uy  = O along the horizontal as shown. It reaches the

target P at vertical depth OA, in the coordinate system as shown.

OA = y = 490 m

From second equation of motion,

y = uy t +1/2 gt²

= 1/2 gt^2

490 = 1/2 × 9.8 t²

t = √100 = 10s

(ii) Distance of the target from the hill is given by

AP = x = Horizontal velocity x time

= 98 × 10

= 980 m

Question-34: A body is projected horizontally from the top of a cliff with a velocity of 9.8 m/s. After what time its horizontal and vertical velocities become equal?

Answer-

ux = uy

9.8 = uy +gt

9.8 = 0 +(9.8)t

9.8t =9.8

t=1s

Question-35: Two balls are thrown with the same initial velocity at angles θ and (90° – θ) with the horizontal what will be the ratio of: (i) maximum heights attained and (ii) horizontal ranges.

Answer-

Two balls are thrown with the same initial velocity at angles θ and (90° - θ) with the horizontal what will be the ratio of: (i) maximum heights attained and (ii) horizontal ranges.

hence, h1 : h2 = tan^2 θ : 1.

—:  end of Motion in Plane Numerical on Projectile Motion ISC Class 11 Ch-5 Nootan Solution of Kumar and Mittal Physics Nageen Prakashan :—

Return to :  Nootan Solutions for ISC Physics Class-11 Nageen Prakashan

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