Question

A ball is projected horizontally. After 3 s from projection its velocity becomes 1.25 times of the
velocity of projection. Its velocity of projection is
(A) 10 m/s
(B) 20 m/s
(C) 30 m/s
(D) 40 m/s​

Answer 1

Answer:

option D

40 m/s

Explanation:

Suppose initial horizontal velocity = x m/s

Since the ball is projected horizontally so velocity's x-component will remain same after 3 sec and y-component will be equal to zero initially but it will change linearly as following:

Equation 1

v[y] =u + at

here a is acceleration due to gravity = 9.81m/s²

vy=(9. 81 m/s²) (3s) = 29.43m/s.

Now that we have both components we can calculate the velocity after 3 seconds with Pythagoras:

v=sqrt(vx²+vy²) = sqrt(x²+29.43²) m/s

Given in the question that after 3s from projection its velocity becomes 1.25 times of the velocity of projection

Equation 2

1.25x = √ (x²+29.43²)

1.5625x² = x² + 29.43²

0.5625x² = 29.43²

x² = 29.43²/0.5625

x² = 1539.8

x = √1539.8

x ≈ 40 m/s