A ball is thrown vertically upwards with a velocity of `20 m s^(-1)` from the top of a multistore building of 25 m high. How high will the ball rise? (Take g = `10 m s^(-2)`)

Explanation

Let us take the Y-axis in the vertically upwards direction with zero at the ground.
Now, `v_(0)=+20 ms^(-1), a=-g=-10 ms^(-2), v=0ms^(-1)`
If the ball rises to height y from the point of launch, then using the eqution `v^(2)=v_(0)^(2)+2a(y-y_(0))`
We get `0=(20)^(2)+2(-10)(y-y_(0))`
Solving, we get, `(y-y_(0))=20 m`
There are two ways of solving the problem
Method I - We split two parts the upwards motion (A to B) and the dowanward motion (B to C) and calculate the corresponding time taken `t_(1)` and `t_(2)`.

Since, the velocity at B is zero, we have
`v=v_(0)+g t`
`0=20-10t_(1)`
`t_(1)=2s`
This is the time in in going from A to B. From B or the point of the maximum height, the ball falls freely under the y-direction . We use equation
`y=y_(0)+v_(0)t+(1)/(2)at^(2)`
We have, `y_(0)=45m, y=0, v_(0)=0a=-g=-10 ms^(-2)`
`0=45+(1)/(2)(-10)t_(2)^(2) rArr t_(2)=3 s`
Total time taken by the ball before it hits the ground
`=t_(1)+t_(2)=2s+3s=5s`
Method II- The total time taken can also be calculated by noting 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)`
`therefore 0=25+20t+((1)/(2))(-10)t^(2)rArr 5t^(2)-20t-25 = 0`
Solving this quadratic equation for t, we get
t = 5 s
Note - The second method is better since we do not have to worry about the path of the motion as the motion is under constant acceleration.