A tennis ball is dropped from 1.80 m above the ground. It rebounds to a height of 1.00 m (assume down to be negative). (a) With what velocity does it hit the ground? (b) With what velocity does it leave the ground? (c) If the tennis ball were in contact with the ground for 0.030 s, find its acceleration while touching the ground.
a. V^2 = Vo^2 + 2g*h.
V^2 = 0 + 19.6*1.8 = 35.28
V = 5.94 m/s.
b. V^2 = Vo^2 + 2g*d.
V^2 = 0 + 19.6*1m = 19.6
V = 4.43 m/s. = Velocity with which it
hits gnd. after it bounces. = Velocity
with which it leaves gnd after it bounces.
To solve this problem, we can use the equations of motion for free-falling objects.
(a) To find the velocity with which the tennis ball hits the ground, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity (0 m/s, as the ball is dropped)
a = acceleration due to gravity (-9.8 m/s^2, assuming downward is negative)
s = distance fallen (1.80 m)
Substituting the given values into the equation:
v^2 = 0^2 + 2*(-9.8)*1.80
v^2 = -35.28
Since the velocity cannot be negative, we take the positive square root:
v = √(35.28)
v ≈ 5.94 m/s
Therefore, the velocity with which the tennis ball hits the ground is approximately 5.94 m/s.
(b) To find the velocity with which the tennis ball leaves the ground, we can use the same equation, but with the distance fallen replaced by the distance risen (1.00 m):
v^2 = 0^2 + 2*(-9.8)*(-1.00)
v^2 = 19.6
Taking the positive square root:
v = √(19.6)
v ≈ 4.43 m/s
Therefore, the velocity with which the tennis ball leaves the ground is approximately 4.43 m/s.
(c) To find the acceleration of the tennis ball while touching the ground, we can use the formula:
a = (v-u) / t
Where:
a = acceleration (unknown)
v = final velocity (unknown, assuming it is the same as in part (b))
u = initial velocity (also unknown)
t = time (0.030 s)
Rearranging the equation:
a = (v-u) / t
a = (4.43 - (-5.94)) / 0.030
a = (4.43 + 5.94) / 0.030
a = 342.33 m/s^2
Therefore, the acceleration of the tennis ball while touching the ground is approximately 342.33 m/s^2.
To calculate the velocity of the tennis ball when it hits the ground, we can use the principle of conservation of energy.
(a) With what velocity does it hit the ground?
Step 1: Calculate the potential energy at the starting position.
Potential energy (PE) = mass (m) * acceleration due to gravity (g) * height (h)
The mass of a tennis ball is usually around 0.057 kg, and the height is given as 1.80 m. The acceleration due to gravity is approximately -9.8 m/s^2 (negative due to downward direction).
PE = 0.057 kg * (-9.8 m/s^2) * 1.80 m
PE = -1.001 kg m^2/s^2
Step 2: Calculate the kinetic energy at the lowest point (ground level), assuming all potential energy is converted into kinetic energy.
Kinetic energy (KE) = (1/2) * mass (m) * velocity^2
Since the ball is about to hit the ground, its velocity at the lowest point will be the same as its velocity just before hitting the ground.
KE = -1.001 kg m^2/s^2
Step 3: Equate potential energy and kinetic energy.
PE = KE
-1.001 kg m^2/s^2 = (1/2) * 0.057 kg * velocity^2
-1.001 kg m^2/s^2 = 0.0285 kg * velocity^2
Step 4: Solve for velocity.
velocity^2 = (-1.001 kg m^2/s^2) / (0.0285 kg)
velocity^2 ≈ 35.09 m^2/s^2
Velocity (v) = √(35.09 m^2/s^2)
v ≈ 5.92 m/s
Therefore, the tennis ball hits the ground with a velocity of approximately 5.92 m/s.
(b) With what velocity does it leave the ground?
Since energy is conserved, the velocity with which the ball leaves the ground will be equal to the velocity with which it hit the ground, but in the opposite direction.
Velocity leaving the ground = -(velocity hitting the ground)
Velocity leaving the ground ≈ -(5.92 m/s)
Velocity leaving the ground ≈ -5.92 m/s (negative indicating an upward direction)
Therefore, the tennis ball leaves the ground with a velocity of approximately -5.92 m/s.
(c) If the tennis ball were in contact with the ground for 0.030 s, find its acceleration while touching the ground.
Given that the time of contact with the ground is 0.030 s, we can use the kinematic equation:
Final velocity (vf) = Initial velocity (vi) + (acceleration * time)
The initial velocity is -5.92 m/s (the velocity just before leaving the ground), and the final velocity is 0 m/s (the velocity just after touching the ground and before the rebound). The time of contact is given as 0.030 s.
0 m/s = -5.92 m/s + (acceleration * 0.030 s)
Solving for acceleration:
acceleration * 0.030 s = 5.92 m/s
acceleration ≈ 5.92 m/s / 0.030 s
acceleration ≈ 197.33 m/s^2
Therefore, the acceleration of the tennis ball while touching the ground is approximately 197.33 m/s^2.