#### A ball is thrown vertically upward with a velocity of 20 ms-1 from a top of a multistory building. The height of the point from where the ball is thrown is 25 m from the ground. The height to which the ball rises from the ground and time taken before the ball hits the ground is (g:=10 ms-2)Option: 1 20 m, 2sOption: 2 40m, 3sOption: 3 40m,5sOption: 4 45m,5s

Initially, the ball is thrown upward from the top of the building(point A)

At maximum height (point B) its velocity becomes zero

For  the upward journey  (AB) :

$\\ \text{Initial velocity},\left ( u \right )=20\ m/s\\ \text{Final velocity},\left ( v \right ) =0 \ m/s \text{(At Maximum height)}\\ \text{Acceleration},\left ( a \right )= -g \text{(Due to gravity)}\\ v=u+at\\0=20-10t \\\Rightarrow t=2s\ \ \$

Therefore, it takes 2 sec to reach the maximum height

$\\s=ut+\frac{1}{2}at^{2} \\\\s=20\times 2+\frac{1}{2}\left ( -10 \right )\times 2^{2} \\\\ s=40-20 \\ \\s=h_{AB}=20m$

For journey B to C:

After reaching the maximum point at B from A. the ball will reach eventually to point C

$\\ h_{BC}=h_{AB}+h_{CA} \\h_{BC}=20+25=45m$

$\\ \text{Now, from second equation of motion}\\ s=ut+\frac{1}{2}at^{2}\\ \\ 45=0+\frac{1}{2}\times \left ( 10 \right )\times t^{2} \\ \\45=5t^{2} \\\\\Rightarrow t=3s \\\\ \text{Total time}=t_{AB}+t_{BC}\\ =2+3=5s$