Question

A person is standing on a trolley which moves horizontally with uniform velocity 2îm/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
(C) Om
(D) 0.8mn
(A) 0.6m
(B) 0.2m​

Answer 1

Answer:

0.8 m

Explanation:

velocity of the ball w.r.t. the trolley = 2j m/s

velocity of the trolley w.r.t. the earth = 2i m/s

Therefore, velocity of the ball w.r.t. the earth = velocity of the ball w.r.t. the trolley + velocity of the trolley w.r.t. the earth = (2i + 2j) m/s

Thus the ball will have a horizontal component of velocity 2 m/s and a vertical component of velocity 2 m/s. It is just like projectile motion.

Time in which the vertical displacement of the ball is zero can be found out using the second equation of motion

$0 = 2t - \frac{1}{2} gt^{2}$

Taking $g = 10 m/s^{2}$

we get $t = \frac{2}{5}$

Now the displacement of the ball will be equal to the horizontal distance traveled by it in 2/5 seconds. Since no acceleration is working in the horizontal direction,

the distance traveled in 2/5 seconds = speed x time

                                                              = (2/5) x 2 = 4/5 = 0.8 m