To find the vertical and horizontal components of the ball’s displacement 2.5 seconds after it is thrown, we can use the initial speed and angle to determine the components of the velocity and then use kinematic equations.
Step 1: Resolve the initial velocity into components
The initial speed (\(v_0\)) is 10.0 m/s, and the angle (\(\theta\)) is 37.0°.
Horizontal component of velocity (\(v_{0x}\)): \[ v_{0x} = v_0 \cdot \cos(\theta) = 10.0 \cdot \cos(37.0^\circ) \] Calculating \(\cos(37.0^\circ)\): \[ \cos(37.0^\circ) \approx 0.7986 \] Now calculate \(v_{0x}\): \[ v_{0x} \approx 10.0 \cdot 0.7986 \approx 7.986 , \text{m/s} \]
Vertical component of velocity (\(v_{0y}\)): \[ v_{0y} = v_0 \cdot \sin(\theta) = 10.0 \cdot \sin(37.0^\circ) \] Calculating \(\sin(37.0^\circ)\): \[ \sin(37.0^\circ) \approx 0.6018 \] Now calculate \(v_{0y}\): \[ v_{0y} \approx 10.0 \cdot 0.6018 \approx 6.018 , \text{m/s} \]
Step 2: Horizontal displacement
The horizontal displacement (\(x\)) after time \(t\) can be calculated using: \[ x = v_{0x} \cdot t \] Substituting in the values: \[ x = 7.986 \cdot 2.5 \approx 19.965 , \text{m} \]
Step 3: Vertical displacement
The vertical displacement (\(y\)) can be calculated using the following kinematic equation, considering the effect of gravity (\(g = 9.81 , \text{m/s}^2\)): \[ y = v_{0y} \cdot t - \frac{1}{2} g t^2 \] Substituting in the values: \[ y = 6.018 \cdot 2.5 - \frac{1}{2} \cdot 9.81 \cdot (2.5)^2 \] Calculating: \[ y = 15.045 - \frac{1}{2} \cdot 9.81 \cdot 6.25 \] \[ y = 15.045 - 30.67125 \approx -15.62625 , \text{m} \]
Conclusion
After 2.5 seconds, the vertical and horizontal components of the ball’s displacement are approximately:
- Horizontal Displacement (\(x\)): \(19.97 , \text{m}\)
- Vertical Displacement (\(y\)): \(-15.63 , \text{m}\)
This indicates that the ball has fallen approximately 15.63 meters below its original height.