Question

equation of motion for a projectile launched at an angle of 60 degrees with an initial velocity of 25 m/s

Answer 1

To analyze the motion of a projectile launched at an angle of \( 60^\circ \) with an initial velocity of \( 25 \, \text{m/s} \), we'll break down the initial velocity into its horizontal and vertical components.

1. **Determine the components of initial velocity**:
- Horizontal component (\( V_{0x} \)):
\[
V_{0x} = V_0 \cdot \cos(\theta) = 25 \cdot \cos(60^\circ) = 25 \cdot 0.5 = 12.5 \, \text{m/s}
\]
- Vertical component (\( V_{0y} \)):
\[
V_{0y} = V_0 \cdot \sin(\theta) = 25 \cdot \sin(60^\circ) = 25 \cdot \frac{\sqrt{3}}{2} \approx 21.65 \, \text{m/s}
\]

2. **Equations of motion**:
The projectile's motion can be described using the following equations:

- **Horizontal motion** (constant velocity):
\[
x(t) = V_{0x} \cdot t = 12.5 \cdot t
\]

- **Vertical motion** (constant acceleration due to gravity):
\[
y(t) = V_{0y} \cdot t - \frac{1}{2} g t^2
\]
where \( g \approx 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity. Thus,
\[
y(t) = 21.65 \cdot t - \frac{1}{2} \cdot 9.81 \cdot t^2
\]
Simplifying further:
\[
y(t) = 21.65 \cdot t - 4.905 \cdot t^2
\]

3. **Summary**:
The equations of motion for the projectile are:
- \( x(t) = 12.5 \cdot t \)
- \( y(t) = 21.65 \cdot t - 4.905 \cdot t^2 \)

These equations can be used to predict the position of the projectile at any time \( t \) during its flight.