A ball is thrown vertically upwards with a velocity of `20ms^-1` from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0 m from the ground.
How high will the ball rise?

Explanation

(a) Let us take the y-axis in the vertically upward direction with zero at the ground as shown in Fig.
Now, `v_(0)=+20 ms^(-1), a=-g=-10ms^(-2), v=0`
If the ball rises to height y from the point of launch then using the equation
`v^(2)=v_(0)^(2)+2as " "0=(20)^(2)+2(-10)y`
On solving, we get y=20m
(b) The total time taken can also be calculated by nothing the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation.
`y=y_(0)+v_(0)t+1/2 at^(2)" Now "y_(0)=25m y=0m"`
`v_(0)=20ms^(-1) a=-10ms^(-2), t=?`
`v_(0)=20ms^(-1), a=-10ms^(-2), t=?`
Or `5t^(2)-20t-25=0`
Solving this quadratic equation for t, we get t=5s