A baseball thrown from the outfield is released from shoulder height at an initial velocity of 39.2 m/s at an initial angle of 30.0° with respect to the horizontal. What is the maximum vertical displacement that the ball reaches during its trajectory?

Answer 1

To find the maximum vertical displacement, we can use the kinematic equation for vertical motion:

\[y = y_0 + v_{0y}t - \frac{1}{2}gt^2\]

where:
- \(y\) is the vertical displacement
- \(y_0\) is the initial vertical position (assumed to be 0)
- \(v_{0y}\) is the initial vertical velocity
- \(g\) is the acceleration due to gravity (-9.81 m/s^2)
- \(t\) is the time it takes for the ball to reach the maximum height

From the initial velocity in the vertical direction (which is \(v_{0y} = v_0 \cdot \sin(\theta)\)), we can calculate the initial vertical velocity:

\(v_{0y} = 39.2 \cdot \sin(30.0°) = 19.6 m/s\)

At the maximum height, the vertical velocity is momentarily zero. We can use this information to find the time it takes to reach the maximum height:

\(v_{y} = v_{0y} - gt\)

\(0 = 19.6 - 9.81 \cdot t\)

\(t = \frac{19.6}{9.81} = 2\) seconds

Now, we can find the maximum vertical displacement by substituting the time back into the vertical displacement equation:

\[y = 0 + 19.6 \cdot 2 - \frac{1}{2} \cdot 9.81 \cdot 2^2\]
\[y = 39.2 - 19.62 = 19.6\]

Therefore, the maximum vertical displacement that the ball reaches during its trajectory is 19.6 meters.