Question

A ball is projected upwards from the top of the tower with a velocity $50$ m/s making an angel ${30}^{\circ }$ with the horizontal. The height of the tower is $70$m. After how many seconds from the instant of throwing will the ball reach the ground?

• A. $2s$
• B. $5s$
• C. $7s$
• D. $9s$

Answer 1

Answer: (C)

$7s$

Given: initial velocity (v) $=50m/s$

height of the tower (h) $=70m$

angle with the horizontal $\left(\theta \right)={30}^o$

we take $g=10m/se{c}^2$

First ball goes from point $A$ to $B$.

$\therefore$ for projectile motion $AB$ is

$T_1=\frac{2u\sin \theta }{g}=\frac{2\times 50\times \sin {30}^o}{10}$

$T_1=\frac{2\times 50\times \frac{1}{2}}{10}$

$T_1=5\sec$

Now time taken for covering distance between

point $B$ to $C$ is $T_2$

$⟹h=u{T}_2+\frac{1}{2}g{T}_2^2$

$70=u\sin {30}^o{T}_2+\frac{1}{2}\times 10T_2^2$

$70=50\times \frac{1}{2}{T}_2+5T_2^2$

$70=25T_2+5T_2^2$

$⟹5T_2^2+25T_2-70=0$

$⟹T_2^2+5T_2-14=0$

$⟹(T_2-2)(T_2+7)=0$

$⟹T_2=2$ and $T_2=-7$ (not possible)

$\therefore$ Total time $T=T_1+T_2=5+2=7\sec$