Question

A ball is thrown vertically upwards with an initial velocity of 5 m/s from point P. Q is a point 10m vertically below the pount P. calculate the speed of the ball at the point Q where g=10m/s^2​(a) 7.5m/s (b) 10 m/s (c) 15m/s (d) 17.5 m/s

Answer 1

Answer:

Explanation:

First we have to find height attained by the ball when it is thrown up. For this use the formula

Maximum height =u^2/2g

=> 25/20

=> 5/4

Now add 5/4 and 10 to find total height from maximum height to point Q.

5/4+10= 45/4

Distance is equal to 45/4

Now use formula

V^2-U^2=2as

Here u is zero because the ball fractionally stops before coming down.

Therefore the formula is

V^2= 2as

S is equal to 45/ 4

A equal to gravity which is equal to 10 m/s Square

V^2= 0.5x10x45/4

V^2=225

V=225^.5

V=15 m/s

Therefore final velocity is 15m/s