Question

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

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

Answer 1

The correct option is B $2sec$

$u=52m/s$

$\tan x=\frac{5}{12}$

Initial vertical

Velocity $u_y=u\sin \alpha$

$n=15$ $n=u_{y}t=\frac{1}{2}9t^2$

we get $15=\left(52\sin \alpha \right)t-5t^2$

$5t^2-\left(52\sin \alpha \right)t+15=0$ ...(1)

It will give two values of $t$ i.e. $t_1\&t_2$

$t_1<t_2,t_1=$ first time when it goes to $n=15m$ i.e. Point $A$

$t_2$ again $\left(atB\right)$

So we need $t_2-t_1=\sqrt{(t_2+t_1{)}^2-4t_1{t}_2}$

Sum of roots $=t_2+t_2=\frac{52\sin \alpha }{5}$

$t_1{t}_2=\frac{+15}{5}=3$

So we get $t_2-t_2=\sqrt{\frac{(52{)}^2{\sin }^2\alpha }{25}-12}$

as $\tan \alpha =\frac{5}{12}⟹\sin \alpha =\frac{5}{13}$

$⟹t_2-t_1=\sqrt{\frac{(52{)}^2}{{13}^2}-12}=\sqrt{4}=2$