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 desplacement berween `0.5 m` above the ground.

Explanation

a. `H=20 =(u^(2))/(2g) rArr u=sqrt(2gxx20) =20 m s^(-1)`
b. `t_(a) =(u)/(g) =(20)/(10) =2 s`
c. On hitting the ground, the velocity will be equal to the initial velocity but in the opposite direction. Hence, answer is `20 m s^(-1)` .
d. `S_(1)=20xx 0.5-(1)/(2) 10(0.5)^(2), S_(2) =20xx2.5-(1)/(2) 10 (2.5)^(2)`
Required displacement`=S_(2)-S_(1)=10 m`
e. `15 =20xx20t-(1)/(2) 10t^(2) rArr t^(2)-4t + 3=0 rArr t=1 s,3 s`.