Question

A ball is projected upwards from the top of a tower with a velocity of $50{\text{ms}}^{-1}$ making an angle of ${60}^{\circ }$ with the vertical. If the height of the tower is $70\text{m}$, then the ball will hit the ground in

• A. $2\text{s}$
• B. $3\text{s}$
• C. $5\text{s}$
• D. $7\text{s}$

Answer 1

The correct option is D $7\text{s}$

Vertical component of velocity,
$u_y=50\cos \theta {\text{ms}}^{-1}=25{\text{ms}}^{-1}$
(take upward direction as positive)
$g=-10{\text{ms}}^{-2}$
Vertical displacement of the ball from the top of the tower when it reaches ground $=S=-70\text{m}$
Using $S=ut+\frac{1}{2}a{t}^2,$we get
$-70=25t-5t^2\Rightarrow 5t^2-25t-70=0\Rightarrow t^2-5t-14=0\Rightarrow t^2-7t+2t-14=0\Rightarrow t(t-7)+2(t-7)=0\Rightarrow (t-7)(t+2)=0\Rightarrow t=7s$
[rejecting - ve value of $t$]