A tennis ball is dropped from 1.47 m above
the ground. It rebounds to a height of 1 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s
2
. (Let
down be negative.)
Answer in units of m/s.
021 (part 2 of 3) 10.0 points
With what velocity does it leave the ground?
Answer in units of m/s.
022 (part 3 of 3) 10.0 points
If the tennis ball were in contact with the
ground for 0.00728 s, find the acceleration
given to the tennis ball by the ground.
Answer in units of m/s
2
.
V at ground = sqrt(2 g h)
g * Height attained = (1/2) v^2
change in momentum = m * sum of velocity down and velocity up
force = m a = change in momentum / change in time = m (|v|+ |V|) / .00728
so
a = the sum of the absolute values of velocity down and up / .00728
To find the velocity with which the tennis ball hits the ground, we can use the concept of conservation of mechanical energy. The mechanical energy of the ball is conserved if we ignore the effects of air resistance.
Step 1: Calculate the potential energy at the starting height (1.47 m):
Potential Energy = mass * acceleration due to gravity * height
Potential Energy = m * g * h
Potential Energy = m * 9.8 m/s^2 * 1.47 m
Step 2: Calculate the velocity at the rebound height (1 m):
Since the ball rebounds to a height of 1 m, its potential energy at this height will be equal to the initial potential energy:
Potential Energy = m * g * h
Potential Energy = m * 9.8 m/s^2 * 1 m
Step 3: Equate the potential energies at the starting and rebound heights to find the velocity with which the ball hits the ground:
m * 9.8 m/s^2 * 1.47 m = m * 9.8 m/s^2 * 1 m
Step 4: Solve for the velocity with which the ball hits the ground:
Velocity = sqrt(2 * acceleration due to gravity * (height_rebound - height_start))
Velocity = sqrt(2 * 9.8 m/s^2 * (1 m - 1.47 m))
Therefore, the velocity with which the tennis ball hits the ground is -3.84 m/s (negative since down is considered negative).
To find the velocity with which the ball leaves the ground, we can use the concept of conservation of mechanical energy again.
Step 1: Calculate the potential energy at the rebound height (1 m):
Potential Energy = m * g * h
Potential Energy = m * 9.8 m/s^2 * 1 m
Step 2: Calculate the kinetic energy at the rebound height:
Since the ball rebounded and reached a certain height, it release all potential energy and converted it into kinetic energy.
Kinetic Energy = 1/2 * m * (Velocity)^2
1/2 * m * (Velocity)^2 = m * 9.8 m/s^2 * 1 m
Step 3: Solve for the velocity with which the ball leaves the ground:
Velocity = sqrt(2 * acceleration due to gravity * height_rebound)
Velocity = sqrt(2 * 9.8 m/s^2 * 1 m)
Therefore, the velocity with which the tennis ball leaves the ground is 3.13 m/s.
Finally, to find the acceleration given to the tennis ball by the ground, we can use the concept of average acceleration.
Step 1: Calculate the average acceleration:
Acceleration = Change in velocity / Time
Acceleration = (Final Velocity - Initial Velocity) / Time
Acceleration = (0 m/s - 3.13 m/s) / 0.00728 s
Therefore, the acceleration given to the tennis ball by the ground is approximately -430 m/s^2 (negative since the ball is decelerating).
To find the velocity with which the tennis ball hits the ground, we can use the equation for free-fall motion:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
In this case, the initial velocity u is 0 m/s (since the ball was dropped) and the distance traveled s is the total distance the ball fell and rebounded, which is 1.47 m + 1 m = 2.47 m. Additionally, the acceleration a is due to gravity and is -9.8 m/s^2 (negative because it acts in the opposite direction of the positive direction chosen).
Plugging these values into the equation, we have:
v^2 = (0)^2 + 2(-9.8)(2.47)
v^2 = -48.35
v ≈ -6.95 m/s
Therefore, the velocity with which the tennis ball hits the ground is approximately -6.95 m/s. The negative sign indicates that the velocity is directed downwards.
To find the velocity with which the tennis ball leaves the ground, we can use the principle of conservation of energy. The potential energy when the ball is at its maximum height is equal to the kinetic energy just before it hits the ground.
Since the ball rebounds to a height of 1 m, the potential energy at that height is mgh, where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s^2), and h is the height (1 m). The kinetic energy just before hitting the ground is given by (1/2)mv^2, where v is the velocity.
Setting these equal, we have:
mgh = (1/2)mv^2
Simplifying and rearranging, we find:
v = sqrt(2gh)
Substituting the known values, we have:
v = sqrt(2 * 9.8 * 1)
v = sqrt(19.6)
v ≈ 4.42 m/s
Therefore, the velocity with which the tennis ball leaves the ground is approximately 4.42 m/s.
Lastly, to find the acceleration given to the tennis ball by the ground, we can use the equation:
a = (v-u) / t
where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time.
In this case, the initial velocity u is 0 m/s (since the ball was at rest just before contact with the ground), the final velocity v is the velocity with which the ball leaves the ground (4.42 m/s), and the time t is given as 0.00728 s.
Plugging these values into the equation, we have:
a = (4.42 - 0) / 0.00728
a ≈ 607.14 m/s^2
Therefore, the acceleration given to the tennis ball by the ground is approximately 607.14 m/s^2.
1. V^2 = Vo^2+2g*h = 0+19.6*1.47 = 28.8
V = 5.37 m/s.
2. V^2 = Vo^2+2g*h
0 = Vo^2+(-19.6)1
Vo = 4.43 m/s.
3. V = Vo+a*T
4.43 = 5.37+a*0.00728
a =