Question

A boy throws a ball in air in such a manner that when the ball is at its
maximum height he throws another ball. If the balls are thrown after a time
difference of 1s then what will be the height attained by them?


​

Answer 1

u=0 m/s

t=1 second

time of ascend,

t=u/g

1=u/9.8

9.8 m/s\

maximum height reached, h=u^2/2g

h=(9.8)^2/2×9.8= 4.9 m

Answer 2

Answer:

The maximum height reached by the ball, as it is throne by the boy, was approx 4.9m above the ground level.  

Solution:

This is a problem of ascend of the ball, under effect of gravity. Let assume that ball, at its initial stage had velocity of 0

Let the initial velocity be U.  

$U=0 \frac{m}{s}$

Time of ascent is t

t=1 second

Time of ascend is to be found out.  

$t=u g 1=u 9.8 \frac{m}{s}$

So, the maximum height reached is to be solved.  

Maximum height reached:  

$h=u 22 g h=(9.8) 22 \times 9.8 = 4.9 m$