The velocity of the baseball as it leaves the bat is 38.2 m/s. Determine (a) the maximum height the baseball reaches, and (b) the total timke the baseball is in the air. Assume the baseball is caught at the same height above the ground as it was hit.
To determine the maximum height the baseball reaches and the total time it is in the air, we can use the equations of projectile motion.
(a) Determining the maximum height:
In projectile motion, the vertical motion of the baseball is independent of its horizontal motion. Therefore, we can analyze the vertical motion separately. The initial vertical velocity of the baseball is zero since it is hit horizontally, and the final vertical velocity is also zero when the baseball reaches its maximum height. We can use the following equation to find the maximum height (h):
VFy^2 = VIy^2 + 2 * a * h
Since VFy = 0 (final vertical velocity), and VIy = 0 (initial vertical velocity), the equation simplifies to:
0^2 = 0^2 + 2 * a * h
Simplifying further, we have:
0 = 2ah
Solving for h, we get:
h = 0
This means that the maximum height reached by the baseball is zero.
(b) Determining the total time in the air:
To find the total time the baseball is in the air, we need to consider the projectile motion's vertical component. We can use the equation:
h = VIy * t + 0.5 * a * t^2
Since h = 0, we can rewrite the equation as:
0 = VIy * t + 0.5 * a * t^2
Rewriting the equation further, we get:
0.5 * a * t^2 = -VIy * t
Since a is the acceleration due to gravity (approximately -9.8 m/s^2) and VIy is 0, the equation simplifies to:
0.5 * (-9.8) * t^2 = 0
We can now solve for t:
t^2 = 0 / (-4.9)
t^2 = 0
Since t^2 = 0, the total time the baseball is in the air is zero.
Therefore, the maximum height reached by the baseball is zero, and the total time the baseball is in the air is zero.