In the previous problem, find the magnitude and direction of velocity after `1 s, 3 s` and `5 s`. Also find the distance of the ball from the ground at these instances.

Explanation

Continuing from the previous problem,
`v=u-g t=20-10t`
`h=ut-1/2g t^(2)=10t-5t^(2)`
(h is the displacement from the origin)
At `t=1 s`, `v=20-10xx1=10 m//s`, +ve velocity, i.e. ball is moving in upward direction.
At `h=20xx1-5xx1^2=15m`, +ve displacement, i.e. ball is `15 m` above the origin (top of the tower).
distance of ball above the ground `=60+15=75 m` (Height of tower= 60 m)
At `t=3 s, v=20-10xx3=-10 m//s`, -ve velocity, i.e. ball is moving in downward direction.
At `h=20xx3-5xx3^(2)=15 m`, +ve displacement, i.e. ball is `15 m` above the origin.
Distance of ball above the ground`=60+15=75 m`
At `t=5 s, v=20-10xx5=-30 m//s`, -ve velocity, i.e. ball is moving in downward direction.
At `h=20xx5-5xx5^(6)=-25 m`, -ve displacement, i.e. ball is `25 m` below the origin.
Distance of ball above the ground `=60-25=35 m`