Baseball player A strikes 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 1.2 m above the ground. Player B wishes to duplicate this strike, 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 1.5 m above the ground. What is the magnitude of the initial velocity that player B's ball must be given?

Options

A. 1.3 m/s
B. 1.9 m/s
C. 1.7 m/s
D. 2.1 m/s

Answer

To solve the problem, we need to determine the initial velocity that player B's ball must have in order to travel the same horizontal distance as player A's ball, given that player B strikes the ball from a height of 1.5 m while player A struck it from a height of 1.2 m.

### Step-by-Step Solution:

1. **Determine the Time of Flight for Player A's Ball**:
- Player A strikes the ball at a height of 1.2 m.
- The vertical motion can be described by the second equation of motion:
\[ h = \frac{1}{2} g t^2 \]
where \( h = 1.2 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \).
- Rearranging gives:
\[ t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 1.2}{9.8}} \approx \sqrt{0.2449} \approx 0.4949 \, \text{s} \]

2. **Calculate the Horizontal Distance Traveled by Player A's Ball**:
- The horizontal distance \( d \) traveled by player A's ball can be calculated using:
\[ d = v \cdot t \]
where \( v = 1.9 \, \text{m/s} \) and \( t \approx 0.4949 \, \text{s} \).
- Thus,
\[ d = 1.9 \times 0.4949 \approx 0.9403 \, \text{m} \]

3. **Determine the Time of Flight for Player B's Ball**:
- Player B strikes the ball at a height of 1.5 m.
- Using the same equation of motion:
\[ h = \frac{1}{2} g t^2 \]
where \( h = 1.5 \, \text{m} \).
- Rearranging gives:
\[ t' = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 1.5}{9.8}} \approx \sqrt{0.3061} \approx 0.5532 \, \text{s} \]

4. **Calculate the Required Initial Velocity for Player B's Ball**:
- To travel the same horizontal distance \( d \) in the time \( t' \):
\[ d = u \cdot t' \]
where \( u \) is the initial velocity of player B's ball.
- Rearranging gives:
\[ u = \frac{d}{t'} = \frac{0.9403}{0.5532} \approx 1.698 \, \text{m/s} \]

5. **Final Result**:
- Rounding to two decimal places, the magnitude of the initial velocity that player B's ball must be given is approximately:
\[ u \approx 1.70 \, \text{m/s} \]