To find the magnitude of the acceleration of the ball, we can follow these steps:
1. **Resolve the initial and final velocities into their components.**
2. **Determine the change in velocity.**
3. **Compute the acceleration using the change in velocity and the time of contact.**
4. **Determine the magnitude of the acceleration vector.**
### Step 1: Resolve the velocities into their components
1. **Initial velocity components:**
- Initial speed, \( v_i = 2.11 \, \text{m/s} \)
- Angle, \( \theta_i = -37.0^\circ \)
\[ v_{ix} = v_i \cos(\theta_i) = 2.11 \cos(-37.0^\circ) \]
\[ v_{iy} = v_i \sin(\theta_i) = 2.11 \sin(-37.0^\circ) \]
Using a calculator:
\[ v_{ix} \approx 2.11 \left(\frac{4}{5}\right) = 2.11 \times 0.7986 \approx 1.688 \, \text{m/s} \]
\[ v_{iy} \approx 2.11 \left(-\frac{3}{5}\right) = 2.11 \times (-0.602) \approx -1.272 \, \text{m/s} \]
2. **Final velocity components:**
- Final speed, \( v_f = 3.80 \, \text{m/s} \)
- Angle, \( \theta_f = 150^\circ \)
\[ v_{fx} = v_f \cos(\theta_f) = 3.80 \cos(150^\circ) \]
\[ v_{fy} = v_f \sin(\theta_f) = 3.80 \sin(150^\circ) \]
Using a calculator:
\[ v_{fx} \approx 3.80 \left(-\frac{\sqrt{3}}{2}\right) = 3.80 \times (-0.866) \approx -3.291 \, \text{m/s} \]
\[ v_{fy} \approx 3.80 \left(\frac{1}{2}\right) = 3.80 \times 0.5 \approx 1.900 \, \text{m/s} \]
### Step 2: Determine the change in velocity
\[ \Delta v_x = v_{fx} - v_{ix} = -3.291 - 1.688 \approx -4.979 \, \text{m/s} \]
\[ \Delta v_y = v_{fy} - v_{iy} = 1.900 - (-1.272) \approx 3.172 \, \text{m/s} \]
### Step 3: Compute the acceleration using the change in velocity and the time of contact
\[ \Delta t = 0.19 \, \text{s} \]
\[ a_x = \frac{\Delta v_x}{\Delta t} = \frac{-4.979}{0.19} \approx -26.21 \, \text{m/s}^2 \]
\[ a_y = \frac{\Delta v_y}{\Delta t} = \frac{3.172}{0.19} \approx 16.69 \, \text{m/s}^2 \]
### Step 4: Determine the magnitude of the acceleration vector
\[ a = \sqrt{a_x^2 + a_y^2} \]
\[ a = \sqrt{(-26.21)^2 + (16.69)^2} \approx \sqrt{686.92 + 278.56} \approx \sqrt{965.48} \approx 31.07 \, \text{m/s}^2 \]
So, the magnitude of the acceleration of the ball is:
\[ a \approx 31.07 \, \text{m/s}^2 \]
a ball is moving 2.11 m/s at an angle of -37.0 degrees when it is hit by a racquet. the ball is in contact with the racket 0.19s. after the ball moves with a velocity of 3.80 m/s in a 150 degrees angle. what is the magnitude of the acceleration of the ball?
a=[?] m/s^2
1 answer