Question

A golfer standing on level ground hits a ball with a velocity of $u=52m{s}^{-1}$ at an angle $\alpha$ above the horizontal. If $\tan \alpha =5/12$, then the time for which the ball is at least $15m$ above the ground will be (take $g=10m{s}^{-2})$.

• A. $1s$
• B. $2s$
• C. $3s$
• D. $4s$

Answer 1

The correct option is B $2s$

Let at any time $t$, the ball be at height of $15m$.
$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$\Rightarrow 15=u\sin \alpha t-\frac{1}{2}g{t}^2\Rightarrow 15=52\times \frac{5}{13}t-\frac{1}{2}\times 10t^2$
$\Rightarrow t^2-4t+3=0\Rightarrow (t-1)(t-3)=0$
$\Rightarrow t=1s,t=3s$. Required time is $3-1=2s$.