Question

A person is standing on a trolley which moves horizontally with uniform velocity 2î m/s. At t=0, person throws
a ball with velocity (2j) m/s wrt trolley. Displacement of ball when it returns to initial horizontal level is
(A) 0.6m
(B) 0.2m
(C) Om
(D) 0.8m​

Answer 1

Answer:

The displacement of ball when it returns to initial horizontal level is 0.8 m.

D is correct option.

Explanation:

Given that,

Horizontal velocity = 2i m/s

Vertical velocity = 2j m/s

Horizontal velocity component along x-axis

$V_{x}=V\cos45^{\circ}$

$V=2\times\sqrt{2}$

Vertical velocity component along y -axis

$V=2\times\sqrt{2}$

We need to calculate the displacement of ball

$R =\dfrac{v^2\sin\theta}{g}$

R = range = displacement

V = velocity

g = acceleration due to gravity

Put the value into the formula

$R=\dfrac{(2\sqrt{2})^2\sin90^{\circ}}{9.8}$

$R=\dfrac{8}{9.8}$

$R=0.8\ m$

Hence, The displacement of ball when it returns to initial horizontal level is 0.8 m.