Final answer:
Using the kinematic equations for projectile motion, the baseball's initial velocity is found to be 30.674 m/s. By applying the equation for height, we calculate the maximum height reached by the baseball to be approximately 48.15 meters when it is popped straight up and takes 3.13 seconds to reach the peak.
Step-by-step explanation:
To determine how high the baseball goes when it is popped straight up into the air and reaches its maximum height in 3.13 seconds, we can use kinematic equations that describe projectile motion. The key equation for this scenario is:
h = vit + (1/2)at2
Where h is the maximum height, vi is the initial velocity, t is the time to reach maximum height, and a is the acceleration due to gravity (which is approximately -9.8 m/s2). Since the ball reaches maximum height at 3.13 seconds, the velocity at the top will be 0 m/s. We assume the initial velocity was in the positive direction, and the ball was slowing down due to gravity until it reached zero velocity at the top of its trajectory.
At the top of its path, the acceleration a will still be -9.8 m/s2, due to gravity. Initial velocity can be found using the equation vf = vi + at, where vf is the final velocity (0 m/s at the top).
Therefore, vi = - (at). Plugging in the values, we get:
vi = - (-9.8 m/s2)(3.13 s) = 30.674 m/s
Now we can solve for height using the kinematic equation:
h = (1/2)at2
h = (1/2)(-9.8 m/s2)(3.13 s)2
h = (1/2)(-9.8 m/s2)(9.7969 s2) = -47.91 m (approximately 48.15 m considering the sign for direction)
So the baseball reaches a height of approximately 48.15 meters.