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

Question

A ball is projected from the ground at an angle of $45 °$ with the horizontal surface. It reaches a maximum height of $120 m$ 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 ______ . [2018]

Smart Achievers · Accepted Answer

(30)

Maximum height,

$120 m$

$∴$ $u^{2} = 480 g$

Upon hitting the ground, it loses half of its kinetic energy

$∴$ ${K.E}_{initial} = \frac{1}{2} m u^{2} = 240 m g$

${K.E}_{final} = \frac{1}{2} (240 m g) = 120 m g$

$∴$ $\frac{1}{2} m v^{2} = 120 m g \Rightarrow v^{2} = 240 g$

After the bounce, the maximum height the ball reaches

$∴$ $H' = \frac{v^{2} \sin^{2} θ}{2 g} = \frac{240 g \times (\frac{1}{4})}{2 g} = 30 m$

Solution

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Solution

Ans. (30) Maximum height, H=u2sin2θ2g ⇒ 120=u2(12)2g ∴ u2=480g Upon hitting the ground, it loses half of its kinetic energy ∴ K.Einitial=12mu2=240mg K.Efinal=12(240mg)=120mg ∴ 12mv2=120mg ⇒ v2=240g After the bounce, the maximum height the ball reaches ∴ H'=v2sin2θ2g=240g×(14)2g=30 m

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