Workout the following question
1. A ball of a mass m is dropped from a height h above the ground. Neglecting air resistance,
a. Determine the speed of the ball when it at a height y above the ground
b. Determine the speed of the ball at y ifat the instant of release , it already has an initial upward speed vi at initial altitude h.
a. To determine the speed of the ball when it is at a height y above the ground, we can use the principle of conservation of energy. The initial potential energy of the ball at height h is equal to the sum of its kinetic energy and potential energy at height y.
At height h:
Potential Energy = mgh
Kinetic Energy = 0
Total Energy = mgh
At height y:
Potential Energy = mgh
Kinetic Energy = 1/2 mv^2
Total Energy = mgh + 1/2 mv^2
Setting these two total energies equal gives:
mgh = mgh + 1/2 mv^2
0 = gh + 1/2 v^2
v = sqrt(2gh)
Therefore, the speed of the ball when it is at height y above the ground is sqrt(2gh).
b. To determine the speed of the ball at height y if at the instant of release, it already has an initial upward speed vi at initial altitude h, we need to consider both the initial kinetic energy and potential energy of the ball.
At height h with initial speed vi:
Potential Energy = mgh
Kinetic Energy = 1/2 mvi^2
Total Energy = mgh + 1/2 mvi^2
At height y:
Potential Energy = mgh
Kinetic Energy = 1/2 mv^2
Total Energy = mgh + 1/2 mv^2
Setting these two total energies equal gives:
mgh + 1/2 mvi^2 = mgh + 1/2 mv^2
1/2 mvi^2 = 1/2 mv^2
vi^2 = v^2
vi = v
Therefore, the speed of the ball at height y if it has an initial upward speed vi at initial altitude h is equal to the speed of the ball when it is at height y above the ground.