A ball thrown up from the ground reaches a maximum height of 20 m. Find:
(a)its initial velocity,
(b)the time taken to reach the highest point,
(C ) its velocity just before hittinf the ground,
(d)its displacement between 0.5s and 2.5 s,
(e)time when displacement is 15 m.

Explanation

(a) Using `v^(2)=u^(2)`+2as upward motion,we get `0^(2)=u^(2)+2(-g)(+20)impliesu=19.8 m//s`
(b)`t=(v-u)/(a)=(0-19.8)/(-9.8)=2.02 s`
(c ) For the complete up -down trip.
`v^(2)=u^(2)+2a(0)impliesv^(2)=u^(2)impliesv=-u=-19.8 m//s`
(d)Height at t =0.5s is `y_(1)=19.8(0.5)s` is `y_(1)=19.8(0.5)-4.9(0.5)^(2)=8.68` m.
Height at t=2.5 s is `y_(2)=19.8(2.5)-4.9(2.5)^(2)`=18.9m.
`therefore` Displacement=`y_(2)-y_(1)`
18.9 m-8.68m=+10.2m.
(e)15=19.8t-4.91` t^(2)impliest=1.01s,3.03s`.
At t=1.01 s. ball is going up and at t=3.03s,it is coming down.