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 hitting the ground.
d. Its displacement between `0.5 m` above the ground.

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.