## Question

A rider on an open platform, which is descending at constant speed of 3 ms^{-1}, throws a ball. Relative to platform, ball’s initial velocity is horizontal at 12 m/s. The ground is 10 m below the location where the ball is thrown:

(i):-Where does the ball hit the ground?

(ii):-How long after the ball hits the ground does the platform reach ground level?

(iii):-With what velocity does the ball hit the ground?

### Solution

Due to inertia ball will share the velocity of platform at the instant of projection. Hence horizontal components of ball’s velocity = 12 m/s and vertical components of ball’s velocity = 3 m/s.

Considering the vertical motion,

49*t*^{2} + 30*t* – 100 = 0

Rejecting –ve root, we have

(i):-The distance *AB* = 12 × 1.15 = 13.80 m.

(ii):-The time taken by platform to reach the ground = 10/3 = 3.33 sec. Thus, the time difference = 3.33 – 1.15 = 2.18 s.

(iii):-Let the ball strike the ground with velocity *v*; then horizontal component *v _{h}* = 12 m/s and vertical component

(iv)

#### SIMILAR QUESTIONS

A stone ‘A’ is dropped from the top of a tower 20 m high simultaneously another stone ‘B’ is thrown up from the bottom of the tower so that it can reach just on the top of tower. What is the distance of the stones from the ground while they pass each other.