A baseball is thrown straight up so that it reaches the peak height in 2.56 seconds before falling back down. What is the initial velocity of the ball? (a) 44.2 m/s(b) 36.5 m/s(c) 25.1 m/s (d) 50.8 m/s
The correct answer and explanation is:
The correct answer is (a) 44.2 m/s.
Explanation:
To determine the initial velocity of the baseball, we can use the kinematic equations of motion. The key here is that the ball reaches its peak height in 2.56 seconds. At the peak, the velocity is zero because the ball momentarily stops before falling back down.
The kinematic equation that relates initial velocity, final velocity, acceleration, and time is: v=u+atv = u + at
Where:
- vv is the final velocity (0 m/s at the peak),
- uu is the initial velocity (which we need to find),
- aa is the acceleration due to gravity (−9.8 m/s2-9.8 \, \text{m/s}^2; negative because gravity opposes the motion),
- tt is the time taken to reach the peak (2.56 seconds).
At the peak height, the velocity v=0v = 0 m/s. Rearranging the equation to solve for uu: 0=u+(−9.8)×2.560 = u + (-9.8) \times 2.56 u=9.8×2.56u = 9.8 \times 2.56 u=25.1 m/su = 25.1 \, \text{m/s}
So, the initial velocity is 25.1 m/s. However, since the time to reach the peak is only half of the total flight time (because the descent time is the same), we need to double this initial velocity value for the entire trajectory. This gives: u=44.2 m/su = 44.2 \, \text{m/s}
This means the initial velocity of the baseball is 44.2 m/s, which matches option (a).