A tennis ball isdropped from 1.52m abovethe ground. It rebounds to a height of 0.918 m. With what velocity does it hit the ground? The acceleration of gravity is 9.8 m/s2 . (Let down be negative.) Answer in units of m/s.
(part 2 of 3) With what velocity does it leave the ground? Answer in units of m/s.
(part 3 of 3) If the tennis ball were in contact with the ground for 0.0107 s, find the acceleration given to the tennis ball by the ground. Answer in units of m/s2.
(1/2) m v^2 = m g h
(1/2) v^2 = gh
or
v = sqrt (2 g h) (remember that for dropped object)
so
v at ground = -sqrt (2*9.8*1.52)
now
goes up .918 meter
same deal in reverse
v up at ground = +sqrt (2*9.8*.918)
a = change in velocity/change in time
= [ +sqrt (2*9.8*.918)- - sqrt (2*9.8*1,52) ] / 0.0107
note - - is +
Part 1:
To find the velocity with which the tennis ball hits the ground, we can use the equation:
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.
Since the ball is dropped, the initial velocity is 0 m/s, and the distance traveled is 1.52 m.
Simplifying the equation, we get:
v^2 = 2as
Plugging in the values:
v^2 = 2 * (-9.8 m/s^2) * (-1.52 m)
v^2 = 29.9584 m^2/s^2
Taking the square root of both sides, we get:
v = √(29.9584 m^2/s^2)
v ≈ 5.47 m/s
So, the velocity with which the tennis ball hits the ground is approximately 5.47 m/s.
Part 2:
To find the velocity with which the tennis ball leaves the ground, we can use the fact that the velocity before and after the bounce is the same in magnitude but opposite in direction.
Therefore, the velocity with which the tennis ball leaves the ground is also approximately 5.47 m/s, but in the opposite direction.
Part 3:
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 of contact with the ground.
Since the velocity after the bounce is -5.47 m/s (opposite direction to the drop), the initial velocity is 5.47 m/s, and the time of contact is 0.0107 s, we can plug in the values:
a = (-5.47 m/s - 5.47 m/s) / 0.0107 s
a = -10.94 m/s / 0.0107 s
a ≈ -1022.43 m/s^2
So, the acceleration given to the tennis ball by the ground is approximately -1022.43 m/s^2.
To solve for the velocity with which the tennis ball hits the ground, we can use the equation of motion:
v^2 = u^2 + 2as
where
v = final velocity (unknown)
u = initial velocity (unknown)
a = acceleration due to gravity (-9.8 m/s^2, since it is going downwards)
s = displacement (distance it falls, 1.52 m)
Rearranging the equation, we have:
v = sqrt(u^2 + 2as)
Substituting the given values, we have:
v = sqrt(0^2 + 2*(-9.8 m/s^2)*(1.52 m))
v = sqrt(0 + (-30.288 m^2/s^2))
v = sqrt(-30.288 m^2/s^2)
Since the result is imaginary, it means that the tennis ball does not hit the ground with a real velocity. Instead, it rebounds back.
For part 2, to find the velocity with which the tennis ball leaves the ground, we can use the equation:
v = u + at
where
v = final velocity (unknown)
u = initial velocity (unknown)
a = acceleration due to gravity (-9.8 m/s^2, since it is going upwards)
t = time (unknown)
We can rearrange the equation to solve for u:
u = v - at
Since the final velocity is zero (since the ball reaches its highest point and starts its descent), and the time it takes to reach that point is the same as the time it takes to return to the ground, we can use the rebound time:
u = -9.8 m/s^2 * (0.918 m) / 0.0107 s
u = -811.214 m/s
Therefore, the velocity with which the tennis ball leaves the ground is approximately -811.214 m/s (negative, indicating it is going upwards).
For part 3, to find the acceleration given to the tennis ball by the ground, we can use another equation of motion:
v = u + at
where
v = final velocity (unknown)
u = initial velocity (unknown, assuming it's the same as the value found in part 2, -811.214 m/s)
a = acceleration given by the ground (unknown)
t = time (0.0107 s)
Rearranging the equation, we have:
a = (v - u) / t
Substituting the known values, we have:
a = (0 m/s - (-811.214 m/s)) / 0.0107 s
a = 811.214 m/s / 0.0107 s
a = 75826.168 m/s^2
Therefore, the acceleration given to the tennis ball by the ground is approximately 75826.168 m/s^2.
To find the velocity at which the tennis ball hits the ground, we can use the equation for the final velocity in free fall:
v = √(2gh)
where:
v is the velocity,
g is the acceleration due to gravity (9.8 m/s^2), and
h is the distance dropped (1.52 m).
Plugging in the values, we get:
v = √(2 * 9.8 m/s^2 * 1.52 m)
= √(29.92 m/s^2 * m^2)
= √29.92 m^2/s^2
≈ 5.47 m/s
Therefore, the tennis ball hits the ground with a velocity of approximately 5.47 m/s.
To find the velocity at which it leaves the ground, we can use the equation:
v = √(2gh)
where h is the maximum rebound height (0.918 m). Plugging in the values, we get:
v = √(2 * 9.8 m/s^2 * 0.918 m)
= √(17.9764 m^2/s^2)
≈ 4.24 m/s
Therefore, the tennis ball leaves the ground with a velocity of approximately 4.24 m/s.
To find the acceleration given to the tennis ball by the ground, we can use the equation:
a = (2 * h) / (t^2)
where h is the maximum rebound height (0.918 m) and t is the contact time with the ground (0.0107 s). Plugging in the values, we get:
a = (2 * 0.918 m) / (0.0107 s)^2
= 1.35678 / 0.00011449 s^2
≈ 11856.9 m/s^2
Therefore, the acceleration given to the tennis ball by the ground is approximately 11856.9 m/s^2.