Question

a ball is dropped to three ground from height of 2m The coefficient og restitution is 0.5 To what height will the ball rebound ?

Answer 1

aswer in this pic i hope it helps.

Answer 2

Answer:

The height to which the ball will rebounce = 0.498 m.

Explanation:

The coefficient of restitution is defined as the ratio of the velocity of an object after the collision to the velocity of the object before the collsion.

$\rm Coefficient of restitution, e = \dfrac{Velocity\ just\ after\ collision}{Velocity\ just\ before\ collision}.$

Before collision of the ball with the ground:

Given that the ball is dropped from the height, $\rm h_1 = 2\ m.$

The acceleration that acts on the ball when it is falling is acceleration due to gravity g, whose value is $9.8\ \rm m/s^2.$

Since the ball is dropped, its initial velocity should be zero, $\rm u_1 =0\ m/s.$

Let the final velocity of the ball just before it hits the ground be $\rm v_1$.

Using the following relation,

$\rm v_1^2-u_1^2=2a_1h_1=2gh_1\\v_1^2-0^2=2\times 9.8\times 2=39.2\\v_1^2=39.2\\v_1=6.261\ m/s.$

After collision of the ball with the ground:

After the collision, let the initial velocity of the ball, just after it hits the ground be $\rm u_2$.

At the maximum height $\rm h_2$ to which the ball rebounce, the final speed of the ball will be zero, $\rm v_2 =0\ m/s.$

The acceleration that acts on the ball while bouncing upwards will be -g.

Now, from the definition of the coefficient of restitution,

$\rm e = \dfrac{u_2}{v_1}\\0.5=\dfrac{u_2}{6.261}\\u_2 =6.261\times 0.5=3.13\ m/s.$

Using the following relation,

$\rm v_2^2-u_2^2=2a_2h_2=2(-g)h_2\\0^2-3.13^2=2\times (-9.8)\times h_2\\h_2 = \dfrac{-3.13^2}{-2\times 9.8}=0.498\ m.$