A small pumpkin of mass 2.22 kg is dropped from rest from a height of 2.87 m. The pumpkin rebounds from the ground to a height of 1.12 m.

a. With what speed does the pumpkin hit the ground?
b. With what speed does the pumpkin leave the ground?
c. What impulse was given to the pumpkin by the ground?
I know you use the rules of linear momentum but i just don't understand how to do this, any help please?

(1/2) m v^2 = m g h

so
v1 = -sqrt(2 g h) (coming down)
h = -2.87 coming down

going up same deal but opposite sign for v
v2 = + sqrt(2*9.8*1.12)

change in momentum = m (v2-v1)

To solve this problem, let's break it down into smaller parts and apply the principles of linear momentum.

a. With what speed does the pumpkin hit the ground?

To find the speed at which the pumpkin hits the ground, we can use the principle of conservation of mechanical energy. At the highest point, the pumpkin has potential energy (mgh) and no kinetic energy. When the pumpkin hits the ground, all the potential energy is converted to kinetic energy. We can set up an equation as follows:

Potential Energy at height = Kinetic Energy at ground

mgh = (1/2)mv^2

Where:
m = mass of the pumpkin (2.22 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height of the drop (2.87 m)
v = velocity/speed at ground

Substituting the given values into the equation:

(2.22 kg)(9.8 m/s^2)(2.87 m) = (1/2)(2.22 kg)v^2

Solving for v, we can find the speed at which the pumpkin hits the ground.

b. With what speed does the pumpkin leave the ground?

To find the speed at which the pumpkin leaves the ground, we need to consider the rebound. Since the pumpkin rebounds to a height of 1.12 m, we can use the principle of conservation of mechanical energy once again.

Potential Energy gained during the rebound = Kinetic Energy at the moment of leaving the ground

This time, the rebound is at a lower height, so the potential energy gained will be less. We can set up the equation as follows:

(mgh') / 2 = (1/2)mv'^2

Where:
m = mass of the pumpkin (2.22 kg)
g = acceleration due to gravity (9.8 m/s^2)
h' = height of rebound (1.12 m)
v' = velocity/speed at leaving the ground

Substituting the given values into the equation, we can solve for v' to find the speed at which the pumpkin leaves the ground.

c. What impulse was given to the pumpkin by the ground?

The impulse given to an object is equal to the change in momentum. In this case, the momentum changes due to the rebound from the ground. The momentum before the rebound is -mv (negative because it points downward) and the momentum after the rebound is mv' (since it points upward in the opposite direction).

The impulse can be calculated using the equation:

Impulse = Change in momentum = Final momentum - Initial momentum

Therefore,

Impulse = mv' - (-mv) = mv' + mv

Substituting the given mass and velocity values, we can calculate the impulse given to the pumpkin by the ground.

By applying the principles of conservation of mechanical energy and linear momentum, you can find the answers to your problem. Remember to use the correct units in all calculations to get accurate results.