A ball is thrown vertically upwards with a velocity of `20 ms^(-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 ms^(-2)`.

Explanation

a) Let us take the y-axis in the vertically upward direction with zero at the vertically upward direction with zero at the ground, as shown in fig. 3. 13.
Now `upsilon=+20ms^(-1)`
`" "a=-g=-10m^(-2)`,
`" "upsilon=0ms^(-1)`
If the ball rises to height y from the point of launch, then using the equation
`" "upsilon^(2)+upsilon_(0)""^(2)=2a(y-y_(0))`
We get
`" "0=(20)^(2)+2(-10)(y-y_(0))`
Solving, we get, `(y-y_(0))=20m`
b) We can solve this part of the problem in two ways. Note carefully the methods used.

First method : In the first method, we split the path in two parts : the upward motion (A to B) and the downward motion (B to C) and calculate the corresponding time taken `t_(1)` and `t_(2)`. Since the velocity at B is zero, we have :
`" "upsilon=upsilon_(0)+at`
`" "0=20-10t_(1)`
`" "t_(1)=2s`
Or,
`" "` This is the time in going from A to B. From B, or the point of the maximum height, the ball falls freely under the acceleration due to gravity. The ball is moving in negative y direction. We use equation
`y=y_(0)+upsilon_(0)t+1/2at^(2)`
We have, `y_(0)=45m, y=0, upsilon_(0)=0, a=-g=-10ms^(-2)`
`" "0=45+(1/2)(-10)t_(2)""^(2)`
Solving, we get `t_(2)=3s`
Therefore, the total time taken by the ball before it hits the ground `=t_(1)+t_(2)=2s+3s=5s`.
Second method : 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)+upsilon_(0)t+1/2at^(2)`
Now `y_(0)=25m, y=0m`
`" "upsilon_(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