A ball is thrown vertically upwards with a velocity of $20m{s}^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is 25 m from the ground. How long will it be before the ball hits the ground $(takeg=10m{s}^{-2})?$

There are two ways of solving the problem Method I We split two parts the upward motion (A to B) and the downward motion (B to C) and calculate the corresponding time
taken t1 and t2. Since, the velocity at B is zero, we have 

$v=v_0+gt$

$0=20-10t_1$

$t_1=2s$

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 

$y=y_0+v_{0}t+\frac{1}{2}a{t}^2$

We have, $y_0=45m,y=0,v_0=0,a=-g=-1m{s}^{-2}$

$0=45+\frac{1}{2}(-10)t_2^2$

$\Rightarrow t_2=3s$

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

$y=y_0+v_{0}t+\frac{1}{2}a{t}^2$

Now, $y_0=25m,y=0m,v_0=20m{s}^{-1},a=-10m{s}^{-1}$

$\therefore 0=25+20t+\left(\frac{1}{2}\right)(-10)t^2$

$5t^2-20t-25=0$

Solving this quadratic equation for t, we get

t = 5 s