Question
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?
Answers
GPT 3.5
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.
\[y = y_0 + v_{0y}t - \frac{1}{2}gt^2\]
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.