Question

a ball is projected vertically upward with a velocity u from ground . if t is the time taken by ball to reach the maximum height H , then what will be the height H ,then what will be the height og ball from ground after time 7/5 t ?

Answer 1

Hi champ!!

Well this question is based on kinematic equation

So let's see what is given to us :-
initial velocity =u
time =t
maximum height = H
.We know that at maximum height the value of final velocity V becomes zero

we know that
s=ut+1/2(at^2 )
H=ut+1/2(-gt^2)
H=ut-1/2(gt^2) ...... (1)

we know that
v=u+at
for maximum height
0=u-gt
t=u/g

putting this value in equation (1)
H=(u*u/g)-(gu^2/(2g^2))
H=(u^2/g)-(u^2/(2g))
H=u^2/g(1-1/2)
H=u^2/2g

The distance covered by ball at 7/5t is
s=u(7t/5)-1/2*g*(7t/5)^2
s=7ut/5-1/2*g*49t^2/25
s=7ut/5-(gt^2*49/50)

Hope this helps :)

Answer 2

here is the solution.