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__?
please anyone?​

Answer 1

Answer:

Maximum height reached = 30 m

Step-by-step explanation:

Given:

• The ball is projected at an angle of 45° with the horizontal surface
• Maximum height reached = 120 m
• Upon hitting the ground for the first time it loses half of it's kinetic energy
• After the bounce, the velocity of the ball makes an angle of 30° with the horizontal surface

To Find:

• The maximum height reached after the bounce

Solution:

Before the bounce:

The maximum height attained by a body undergoing projectile motion is given by,

$\sf H_{max}=\dfrac{u^{2} \:sin^{2}\theta }{2g}$

where u is the initial velocity and g is acceleration due to gravity

Substituting the data we get,

$\sf 120=\dfrac{u^{2}sin^{2} 45 }{2g}$

$\sf 120=\dfrac{u^2}{2g} \times \dfrac{1}{2}$

$\sf \dfrac{u^{2} }{4g} =120$

$\sf \dfrac{u^{2} }{g} =480----(1)$

Now by given K.E after the bounce = Half of the initial kinetic energy.

(K.E)' = 1/2 K.E

(K.E)' = 1/2 (1/2 × m u²)

1/2 m v² = 1/2 (1/2 m u²)

v² = u²/2----(2)

Now the height reached after the bounce is given by,

$\sf H_{new}=\dfrac{v^{2}sin^{2} 30 }{2g}$

Substitute the value of v² from equation 2,

$\sf H_{new}=\dfrac{u^{2}}{2}\times \dfrac{1}{2g}\times \dfrac{1}{4}$

$\sf H_{new}=\dfrac{u^{2}}{g}\times \dfrac{1}{16}$

Substitute equation 1 in above equation,

$\sf H_{new}=480\times \dfrac{1}{16}$

$\sf H_{new}=30\:m$

Hence the maximum height reached by the ball after the bounce is 30 m.

Answer 2

Answer 30. 00

Hope it helps all.

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