Question

You release a ball from a height of 1.5 feet. It bounces without losing any energy (whichmeans it will continue bouncing forever and the height of each bounce will be the same).Write an equation which models the height of the ball as a function of time assuming yourelease it at timet= 0 and it takes 2 seconds to strike the ground after you drop it.

Answer 1

`h = cos(\omega )* \sqrt{(2E)/(k) }`

Step-by-step explanation:

Here, we have the equation of motion of the ball is given s follows

h = 0.5·g·t²

Whereby

U = The potential energy = 0.5·k·x² = 0.5·k·A²cos²(ωt )

K = The kinetic energy = 0.5·m·v² = 0.5mω²A²sin²(ωt)

The kinetic energy = 0.5·k·A²cos²(ωt ) as k = mω²

Therefore, the total mechanical energy, E of the ball = K + U

The total mechanical energy, E of the ball =0.5·k·A²(cos²(ωt) + sin²(ωt))

∴ The total mechanical energy, E of the ball = 0.5·k·h² such that the height h is given by the following equation

A =
`\sqrt{(2E)/(k) }`

Where h =A·sin(ωt + ∅) =
`cos(\omega t )* \sqrt{(2E)/(k) }`.

Answer 2

h ( t ) = 0.75*cos ( π*t / 2 ) + 0.75

Step-by-step explanation:

Solution:-

- A ball is dropped from a height of h = 1.5 feet.

- The ball bounces of the ground and back up to the same height h.

- There is no loss of energy in the motion of ball.

- It takes t = 2 seconds for the ball to hit the ground.

- We see that the ball keeps on oscillating up and down to and from same height h.

- We can model the motion of ball as simple harmonic motion. We will take vertical displacement of the ball from ground as "x".

For SHM, the displacement "x" is a function of time as follows:

x = A*cos ( w*t ) + B

Where,

A : The amplitude of the motion about mean displacement

w : The angular frequency of the motion

B : The mean position of the motion.

- The mean position of the ball motion and its amplitude must add up to total initial height ( h ) from which the ball is dropped:

B = h / 2 = 1.5 / 2 = 0.75 feet

A = h / 2 = 1.5 / 2 = 0.75 feet

- The angular frequency (w) of the ball can be determined from:

w = 2π / T

Where T: The time period of motion. ( from h back to h ).

- We know that the ball takes t = 2 seconds from height ( h ) to ground that is half the motion. Hence, the half of time period, 0.5*T = 2 seconds. Hence,

w = 2π / 4 = π / 2

- Then we can express our height "h" as a function of time "t" as:

h ( t ) = 0.75*cos ( π*t / 2 ) + 0.75