A batter hits a pop-up straight up over home plate at an initial velocity of 29 m/s. The ball is caught by the catcher at the same height that it was hit. At what velocity will the ball land in the catcher's mitt?
none of these
29 m/s
14.5 m/s
-14.5 m/s
-29 m/s
-29. Total energy is conserved so the kinetic energy (and v^2) when caught must be the same as when the ball was hit. However, the direction of the velocity changes from up to down.
To determine the velocity at which the ball lands in the catcher's mitt, we need to consider the concept of conservation of energy.
When the ball is hit straight up, it reaches its maximum height and then falls back down. At the maximum height, the ball's velocity is zero, and all of its initial kinetic energy is converted into potential energy. As the ball falls back down, this potential energy is converted back into kinetic energy.
Since the ball is caught at the same height it was hit, this means that the potential energy at the maximum height is equal to the potential energy just before it lands in the catcher's mitt.
The potential energy of an object is given by the equation: PE = m * g * h, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
In this case, since the mass of the ball is not provided, we can ignore it as it cancels out when comparing the potential energies. The acceleration due to gravity, g, is approximately 9.8 m/s^2.
Therefore, the potential energy at the maximum height is equal to the potential energy just before it lands in the catcher's mitt:
PE_maxHeight = PE_landingVelocity
m * g * h_maxHeight = m * g * h_landingVelocity
Since the height is the same in both cases, h_maxHeight = h_landingVelocity, so we can cancel it out:
g * h_maxHeight = g * h_landingVelocity
Now we can solve for the velocity at which the ball lands in the catcher's mitt.
Given that the initial velocity is 29 m/s and the acceleration due to gravity is 9.8 m/s^2, we can use the following equation of motion to find the time it takes for the ball to reach the maximum height:
v_f = v_i + a * t
0 = 29 m/s - 9.8 m/s^2 * t_maxHeight
Solving for t_maxHeight:
t_maxHeight = 29 m/s / 9.8 m/s^2
t_maxHeight ≈ 2.96 s
Now, we can use the same equation to find the time it takes for the ball to land in the catcher's mitt:
v_f = v_i + a * t
0 = v_landing - 9.8 m/s^2 * t_landing
Solving for v_landing:
v_landing = 9.8 m/s^2 * t_landing
Since we know that the time it takes for the ball to reach the maximum height is 2.96 s, we can substitute it into the equation:
v_landing = 9.8 m/s^2 * 2.96 s
v_landing ≈ 29 m/s
Therefore, the velocity at which the ball will land in the catcher's mitt is 29 m/s.
So the answer is: 29 m/s.
To find the final velocity of the ball when it lands in the catcher's mitt, we can use the principle of conservation of energy. Since the ball is caught at the same height it was hit, there is no change in potential energy. Therefore, the final kinetic energy of the ball will be equal to the initial kinetic energy.
The initial kinetic energy can be calculated using the formula:
KE = 0.5 * mass * velocity^2
Since the mass of the ball is not given, we can assume it to be constant. Therefore, we can disregard the mass in our calculations.
The initial kinetic energy is:
KE_initial = 0.5 * velocity_initial^2
The final kinetic energy is:
KE_final = 0.5 * velocity_final^2
Since the initial kinetic energy is equal to the final kinetic energy, we can set the two equations equal to each other:
0.5 * velocity_initial^2 = 0.5 * velocity_final^2
Rearranging the equation to solve for the final velocity, we get:
velocity_final = sqrt(velocity_initial^2)
Substituting the given initial velocity (29 m/s) into the equation, we get:
velocity_final = sqrt(29^2)
velocity_final = sqrt(841)
velocity_final ≈ 29 m/s
Therefore, the ball will land in the catcher's mitt with a velocity of approximately 29 m/s.
So the correct answer is 29 m/s.