Question

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 1

Hint: The ball is thrown making an angle with the horizontal. The path covered by the body is called projectile. Hence this is a problem of projectile motion and we have to use relevant formulas to find out the answers. Also, we can use conservation of energy to find the velocity.

Complete step by step answer:
Angle made with the horizontal is 45°. The maximum height reached is 120 m. the maximum height is given by the formula \[H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]
\[120=\dfrac{{{u}^{2}}{{\sin }^{2}}45}{2\times 9.8}\]
\[\dfrac{{{u}^{2}}}{4g}=120\]-----(1)
When it hits the ground it loses half of its kinetic energy, kinetic energy is given by the formula, \[KE=\dfrac{m{{u}^{2}}}{2}\].
Any decrease will be on account of velocity as mass does not change.
$
KE=\dfrac{m{{u}^{2}}}{2} \\
\implies KE'=\dfrac{KE}{2} \\
\implies \dfrac{m{{u}^{2}}}{2}=\dfrac{m{{v}^{2}}}{4} \\
\implies v=\dfrac{u}{\sqrt{2}} \\
$
Now final height reached after once striking the ground will be, this time the angle with the horizontal is 30°
${{H}_{2}}=\dfrac{{{v}^{2}}{{\sin }^{2}}\theta '}{2g}$
$\implies {{H}_{2}}=\dfrac{{{(\dfrac{u}{\sqrt{2}})}^{2}}{{\sin }^{2}}30}{2g}=\dfrac{{{u}^{2}}}{16g}$
$\implies {{H}_{2}}=\dfrac{{{u}^{2}}}{16g}$
By Using eq (1) we get,
\[{{H}_{2}}=\dfrac{{{u}^{2}}}{4g}\times \dfrac{1}{4}=\dfrac{120}{4}=30\]

So, the value of height achieved is 30 m. So, the maximum height it reaches after the bounce, in metres, is 30.

Note:
A projectile is any object thrown by the exertion of a force. Always in the formula the angle used is made with the vertical and if in the question it is given that angle is made with the vertical then we have to just subtract the given angle from \[{{90}^{0}}\].