Certainly! Let's approach this problem step-by-step.
### Calculation of Velocity after 3 Seconds
1. Given data:
- Initial velocity (\(\(v_0\)\)) = 100 m/s
- Angle of projection (\(\(\theta\)\)) = 30 degrees
- Time (\(\(t\)\)) = 3 seconds
- Acceleration due to gravity (\(\(g\)\)) = 9.8 m/s²
2. Decomposition of the initial velocity:
- Horizontal component (\(\(v_{0x}\)\)):
\(\[
v_{0x} = v_0 \cdot \cos(\theta)
\]\)
\(\[
v_{0x} = 100 \cdot \cos(30^\circ) = 100 \cdot \frac{\sqrt{3}}{2} \approx 86.60 \, \text{m/s}
\]\)
- Vertical component (\(\(v_{0y}\)\)):
\(\[
v_{0y} = v_0 \cdot \sin(\theta)
\]\)
\(\[
v_{0y} = 100 \cdot \sin(30^\circ) = 100 \cdot \frac{1}{2} = 50 \, \text{m/s}
\]\)
3. Velocity components after 3 seconds:
- Horizontal velocity (\(\(v_x\)\)) remains constant as there is no horizontal acceleration:
\(\[
v_x = v_{0x} = 86.60 \, \text{m/s}
\]\)
- Vertical velocity (\(\(v_y\)\)) after time \(\(t\)\):
\(\[
v_y = v_{0y} - g \cdot t
\]\)
\(\[
v_y = 50 - 9.8 \cdot 3 = 50 - 29.4 = 20.6 \, \text{m/s} \, \text{(downward)}
\]\)
4. Magnitude of the velocity vector:
\(\[
v = \sqrt{v_x^2 + v_y^2}
\]\)
\(\[
v = \sqrt{(86.60)^2 + (20.6)^2} \approx \sqrt{7500 + 424.36} \approx \sqrt{7924.36} \approx 89 \, \text{m/s}
\]\)
### Calculation of Time When Acceleration is Perpendicular to Velocity
5. Understanding perpendicularity:
- Acceleration is due to gravity and acts vertically downwards.
- At time \(\(t\)\), velocity will be perpendicular to acceleration if the vertical component of velocity is zero, as the horizontal component is still \(\(v_x\)\).
6. Setting up the condition:
\(\[
v_y = v_{0y} - g \cdot t = 0
\]\)
\(\[
50 - 9.8 \cdot t = 0
\]\)
\(\[
t = \frac{50}{9.8} \approx 5.10 \, \text{seconds}
\]\)
### Summary of Results
- The velocity of the projectile after 3 seconds is approximately 89 m/s.
- The time when its acceleration is perpendicular to its velocity is approximately 5.10 seconds.
