A ball thrown up verically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up. (b) the maximum height it reaches, and (c ) its position after 4 s.

Explanation

Since the ball thrown up vertically returns to the thrower in 6 seconds , this means that the ball will take half of this time that is `=6/2=3` seconds to go from the thrower its maximum height and the thrower. (a) Calculation of velocity with ball was thrown up
Here, Final velocity ,v=0 (The ball stops )
Intial velocity ,u=? (To be calulated )
Acceleration due `g=-9.8 m//s^(2)` (Ball goes up )
to gravity
And Time taken to t=3s
reach the top
Now v=u+gt
So, `0=u+(-9.8)xx3`
0=u-29.4
u=29.4 m/s
Thus the velocity with which of the ball was thrown up is 29.4 meters per second .
(b ) Calcualtion of the maximum height reached we know that : `v^(2)=u^(2)+2gh`
So, `(0)^(2)=(29.4)^(2)+2xx(-9.8)xxh`
0=864.36-19.6 h
19.6 h= 864.36
` h=(864.36)/(19.6)`
h=44.1 m
Thus , the maximum height reached by the ball is 44.1 metres .
(c ) Calulation of position of ball after 4 s
we have seen above that the maximum height in 3 seconds .So all that we have to do is to find the hieght (or distance ) by which the freely falling ball comes from the top in 1 second ( because 3 s+ 1 s=4 s ).
Now, `h=ut+1/2gt^(2)`
So, h=0xx1+1/2xx9.8xx(1)^(2)` (because t=1s)
h=0+4.9xx1
h=4.9 m
Thus ,the positon of the ball after 4 seconds of throwing is 4.9 metres below the maximum height reached .