Question

A rock is shot with a sling shot at angles 30, 45, 60 and 90.

• which launch angle will the rock's hang time be the greatest

• which launch angle will the rock's horizontal and vertical displacement will be the greatest

Answer 1

Initial horizontal velocity of rock , vertical velocity of rock $u_y=u\sin\theta$

Using equation $v=u+at$

Final horizontal velocity when the rock reaches maximum height is $v_x=u_x=u\sin\theta$ ,final vertital velocity at maximum height is $v_y=0$

Time taken to reach maximum height is .

$v_y=u_y-gt'$

   $\Rightarrow t'=\frac{u_y}{g}=\frac{u\sin\theta}{g}$

Time taken to reach the ground is double the time taken to reach maximum height , $t=2 t'=\frac{2u\sin\theta}{g}$

For the given angles, $\sin\theta$ will be maximum when $\theta=90\degree$

Hence rock's hang time will be greatest when $\theta=90\degree$

------------------------------

Using $s=ut+\frac{1}{2}at^2$

Rock's maximum vertical displacement $y=u\sin\theta-\frac{1}{2}gt'^2$ ( time taken to reach maximum vertical displacement is )

$y=u\sin\theta*\frac{u\sin\theta}{g}-\frac{1}{2}g\left ( \frac{u\sin\theta}{g} \right )^2=\frac{u^2\sin^2\theta}{2g}$

for maximum vertical displacement (y) , $\sin^2\theta=1$ $\Rightarrow \theta=90\degree$

Rock's maximum horizontal displacement is ( is total time taken to reach ground)

$x=u\cos\theta t=u\cos\theta*\frac{2u\sin\theta}{g}=\frac{u^2 \sin2\theta}{g}$

For maximum horzontal displacement (x)  ,   $\sin2\theta=1$    $\Rightarrow \theta=45\degree$