Question

A ball is thrown from the top of a building 45 m high with a speed 20 m s${}^{-1}$ above the horizontal at an angle of ${30}^o$. Find
a. The time taken by the ball to reach the ground.
b. The speed of ball just before it touches the ground.

Answer 1

Given $v=20m{s}^{-1},\theta ={30}^o,H=45m$.

a. As the ball has been projected at an angle of ${30}^o$ above horizontal, so first of all we need to analyse the velocity horizontally and vertically. This will be useful while using distance-time relation in horizontal and vertical directions.

$v_{xi}=vcos{30}^o=20\times \frac{\sqrt{3}}{2}=10\sqrt{3}m{s}^{-1}$

$v_{yi}=vsin{30}^o=20\times \frac{1}{2}=10m{s}^{-1}$

It will be easy for us to use distance-time relation in vertical as it will involve less calculation.

In y-direction: $-45=10t-\frac{1}{2}\times g{t}^2\Rightarrow t^2-2t=0$

which on solving gives $t=1+\sqrt{1}0s$ (positive value), (other value is $1-\sqrt{1}0s$, a negative value of time is not acceptable).

b. $v_{yf}=10-10\times (1+\sqrt{1}0)=-10\sqrt{1}0m{s}^{-1}$

$v_f=\sqrt{v_{yf}^2+v_{xf}^2}$

$=\sqrt{(10\sqrt{1}0{)}^2+(10\sqrt{3}{)}^2}$

$=10\sqrt{3}m{s}^{-1}$