Question

A ball is projected vertically down with an initial velocity from a height of 20m on to a horizontal floor. during the impact it loses 50% of its energy and rebounds to the same height. the velocity of projection is

Answer 1

Answer:

v0=19.82ms

Explanation:

v = v0 + g t
h = v0 t + g t²/2 = 20
(Everything measured downwards)

Ek = (1/2) m v²
=> Ep = m g h = 196.2 m = (1/4) m v²
=> v ² = 784.8
=> v = 28.014 m/s

28.014 = v0 + 9.81 t
20 = v0 t + 4.905 t²

⇒ 20 = (28.014 - 9.81 t) t + 4.905 t²
= 28.014 t - 4.905 t²
⇒ 4.905 t² - 28.014 t + 20 = 0
This is a quadratic equation in t
t = (28.014 ± 19.809)/9.81
The solution with the - sign yields
t = 0.836 s ⇒ v0 = 19.82 m/s

Answer 2

Answer:

The velocity of the projection is 20m/s.

Explanation:

The energy at the bottom is calculated as,

$E=mgh+\frac{1}{2} mv^2$       (1)

Where,

E=energy at the bottom

m=mass of the body which is projected

g=acceleration due to gravity=10m/s²

h=height from which the body is projected

v=velocity of the body

From the question we have,

Height(h)=20m

By placing the required values in equation (1) we get;

$E=m\times 10\times 20+\frac{1}{2} mv^2$      (2)

Since there is a loss of 50% of energy when the ball rebounds to the same height. So,

$\frac{1}{2} (m\times 10\times 20+\frac{1}{2} mv^2)=mgh$

$100m+\frac{1}{4}mv^2 =m\times 10\times 20$

$\frac{1}{4}mv^2 =200m-100m$

$\frac{1}{4}mv^2 =100m$

$v^2=400$

$v=\sqrt{400}$

$v=20m/s$

Hence,  the velocity of the projection is 20m/s.

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