Question

You and your best friend are bored on a saturday afternoon and decide to measure the impulse acting on a basketball when it bounces off the ground. the ball with a mass of m is dropped from rest at a height h. assume no fritcional forces are in play such that the ball bounces as a perfectly elastic colision.
derive an experession for the velocity of the ball jus before it collides with the ground.

Answer 1

Answer:

v = √(2gh)

Assuming g = 9.81m/s^2

v = √(19.62h)

Explanation:

Let v represent the speed of the ball just before it collide with the ground.

Applying the equation of motion;

v^2 = u^2 + 2as

Where;

v = final velocity

u = initial speed = 0 (starting from rest)

a = acceleration= g acceleration due to gravity

s = distance covered = h

So, substituting the values;

v^2 = 0^2 + 2gh

v^2 = 2gh

v = √(2gh)

Assuming g = 9.81m/s^2

v = √(19.62h)

Answer 2

Answer:

V = (19.62h)^0.5

Explanation:

We can apply the principle of conservation of momentum in this case.

The potential energy due to the height of fall is transformed into the kinetic energy of the ball as it falls.

PE = mgh

Where m is the mass of the ball,

g is the acceleration due to gravity = 9.81 m/s2,

h is the height of fall.

The kinetic energy of the ball = 1/2(mv^2)

Where c is the velocity of the ball.

Equating both energy,

mgh = 1/2(mv^2)

2gh = v^2

v = (2gh)^0.5 = (19.62h)^0.5

Or velocity V is equal to the square root of (19.62h)