Question

A boy throws a ball vertically up . it returns to the ground after 5 seconds . find : i. the maximum height reached by the ball. ii. the velocity with which the ball is thrown up.

Answer 1

Answer:

The equation is given as v  

2

−u  

2

=2as.

where,

v = final velocity,  

u = initial velocity,  

a = acceleration due to gravity, 10m/s  

2

,  

s = h = height traveled.

Therefore, we can find u with above info. In this case, v = 0 at the top of the path.

0−u  

2

=2×10×5=10m/s.

Hence, the initial velocity of the ball was 10 m/s.

Answer 2

It would have gone up in 2.5 sec and come down in 2.5 s

s = ut + 1/2 at²

s = 0 = 1/2 (10) (2.5)²

s = 31.25

v = 0 + (10)(2.5) = 25