Discussion Forum

A ball is projected from the ground at an angle of 45° with the horizontal surface . It reaches a maximum  height of 120m and  returns to the ground. Upon hitting the ground for the first time, it loses  half of its kinetic energy . Immediately after the bounce , the velocity of the ball makes an angle of 30° with the horizontal surface . The maximum height it reaches after the bounce, in metres, is.........

Answer:

Explanation:

$H=\frac{u^{2}\sin^{2}45^{0}}{2g}=120m$

$\Rightarrow$   $\frac{u^{2}}{4g}=120m$

  If speed  is v after the first collision, then speed should remain $\frac{1}{\sqrt{2}}$  times, as kinetic energy has reduced to half.

$\Rightarrow$    $v=\frac{u}{\sqrt{2}}$

$\therefore$     $h_{max}= \frac{v^{2}\sin^{2}30^{0}}{2g}$

                                       = $ \frac{(\frac{u}{\sqrt{2}})^{2}\sin^{2}30^{0}}{2g}$

                                      = $(\frac{u^{2}/4g}{4})=\frac{120}{4}=30m$