Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.9 m/s parallel to the ground. Upon contact with the bat, the ball is 1.5 m above the ground. Payer B wishes to duplicate this bunt, in so far as he also wants to give the ball a velocity parallel to the ground and have his ball travel the same horizontal distance as player A's ball does. However, player B hits the ball when it is 2 m above the ground. What is the magnitude of the initial velocity that player B's ball must be given?
To find the magnitude of player B's initial velocity, we can use the principle of conservation of energy. The potential energy at the start of the bunt is converted into kinetic energy at the moment of contact with the bat. Since both players hit the ball with the same horizontal velocity, we can assume that the change in horizontal kinetic energy is the same for both.
Let's break down the problem step by step:
1. Calculate the initial potential energy for player A's bunt:
- Mass of the baseball (m) is not given, but its weight (mg) cancels out in this calculation.
- Height of the ball above the ground (h) = 1.5 m
- Acceleration due to gravity (g) = 9.8 m/s^2
- Potential energy (PE) = mgh
2. Calculate the initial potential energy for player B's bunt:
- Height of the ball above the ground (h) = 2 m
- Potential energy (PE) = mgh
3. Set the potential energy of player B's bunt equal to the potential energy of player A's bunt and solve for the initial velocity (v):
- mgh_playerB = mgh_playerA
- The mass of the ball (m) cancels out.
- gh_playerB = gh_playerA
- v_playerB = √(2gh_playerA)
4. Plug in the given values:
- g = 9.8 m/s^2
- h_playerA = 1.5 m
5. Calculate the magnitude of player B's initial velocity:
- v_playerB = √(2 * 9.8 * 1.5)
By calculating this expression, you will get the final answer for player B's magnitude of the initial velocity.