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 of ball from ground after time 7\5t?

Answer 1

, v = u + at. Hence, we can write t = (v-u)/a

Also, we know that, Distance = average velocity × Time

Therefore, for constant acceleration we can write: Average velocity = (final velocity + initial velocty)/2 = (v+u)/2

Hence, Distance (s) = [(v+u)/2] × [(v-u)/a]

or s = (v² – u²)/2a

or 2as = v² – u²

or v² = u² + 2as

Answer 2

s = h , v = 0 , initial velocity = u , time of journey = t
the height in (7/5)t
since the 7/5 = 1.4
and it is greater than 1 hence the height in (1.4)t
u = 0 , s = ? time = 1.4t
s = ut + (at^2)/2
= (10)(1.4)(1.4)(t)/2
= 28t ....