A ball is thrown vertically upwards with a velocity of `20 m s^(-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. (a) How high will the ball rise ? and (b) how long will it be before the ball hits the ground? Take `g=10 m s^(-2)`

Explanation

Let us take the y-axis in the vertically upward direction with zero at the ground, as shown in Fig.
Now `v_(0)=+20ms^(-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=?`
`0=25+20t+(1//2)(-10)t^(2)`
Or, `5t^(2)-20t-25=0`
Solving this quadratic equation for t, we get `t=5s`