Question

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

Answer 1

Given $\tan a=\frac{5}{12}\Rightarrow \sin a=\frac{5}{\sqrt{{12}^2+5^2}}=\frac{5}{13}$
Let at any time $t$, the ball is at height of $15\text{m}$.
$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$\Rightarrow 15=4\sin \theta t-\frac{1}{2}g{t}^2$
$\Rightarrow 15=52\times \frac{5}{13}t-\frac{1}{2}\times 10t^2$
$\Rightarrow t^2-4t+3=0\Rightarrow (t-1)(t-3)=0$
$\Rightarrow t=1\text{s},t=3\text{s}$,

So, ball is at a height of 15 m at $t=1$( while moving up) and $t=3$ (while moving down).
Hence ball is at a height greater than 15 m , in the time interval between these instants.
required time: $3-1=2\text{s}$.