Question

Paragraph
A small ball is thrown with initial speed u at an angle theta (with the horizontal) from a smooth horizontal surface. The ball
makes several impacts with the surface. The coefficient of restitution is e.
Q)The angle which the resultant velocity makes with horizontal, just after 4th impact (collision)
(1)tan-1(e4 tanø)
and other options,the answer is 1 but i need explanation,plz no spam ,
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Answer 1

✪ᴀɴsᴡᴇʀ✪

• The resultant velocity makes an angle $\bf{tan^{-1}[e^4tan(\theta)]}$ with the horizontal after fourth collision.

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Let

• $\bf{u_x, u_y}$ be the horizontal and vertical component of initial velocity.
• $\bf{\phi }$ be angle between resultant velocity and horizontal after fourth impact.

Then,

$\to u_x = ucos(\theta)$

$\to u_y = usin(\theta)$

On impact with the ground,

• Horizonal Component of velocity remains unchanged.

• Vertical Component of velocity changes by a factor of 'e'.

So, after first impact,

$\to u_{x1} = ucos(\theta)$

$\to u_{y1} = uesin(\theta)$

After each collision the velocity will changes as explained above. Continuing in this manner, after fourth impact,

$\to u_{x4} = ucos(\theta)$

$\to u_{y4} = ue^4sin(\theta)$

Now,

$\:\:\:\:\:\:\: tan(\phi) = \frac{u_{y4}}{u_{x4} }$

$\implies tan(\phi) = \frac{ue^4sin(\theta)}{ucos(\theta)}$

$\implies tan(\phi) = e^4\frac{sin(\theta)}{cos(\theta)}$

$\implies tan(\phi) = e^4tan(\theta)$

$\implies \boxed{ \phi = tan^{-1}[e^4tan(\theta)]}$

Answer 2

Answer:

Tq for answer bro

Explanation:

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