Question

What is the trajectory of a projectile launched at an angle of 60 degrees with an initial velocity of 25m/s?

Answer 1

To determine the trajectory of a projectile launched at an angle \( \theta = 60^\circ \) with an initial velocity \( v_0 = 25 \, \text{m/s} \), we can break down the motion into its horizontal and vertical components.

1. **Resolve the initial velocity:**
- The horizontal component of the initial velocity (\( v_{0x} \)) is given by:
\[
v_{0x} = v_0 \cdot \cos(\theta) = 25 \cdot \cos(60^\circ) = 25 \cdot 0.5 = 12.5 \, \text{m/s}
\]
- The vertical component of the initial velocity (\( v_{0y} \)) is given by:
\[
v_{0y} = v_0 \cdot \sin(\theta) = 25 \cdot \sin(60^\circ) = 25 \cdot \frac{\sqrt{3}}{2} \approx 21.65 \, \text{m/s}
\]

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

- Horizontal motion:
\[
x(t) = v_{0x} \cdot t = 12.5 \, t
\]

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

3. **Time of flight:**
The time of flight (\( T \)) can be found by determining when the projectile returns to \( y = 0 \) after launch. Setting the vertical motion equation equal to zero gives:
\[
21.65t - 4.905t^2 = 0
\]
Factoring out \( t \):
\[
t(21.65 - 4.905t) = 0
\]
This gives \( t = 0 \) (launch), and:
\[
21.65 - 4.905t = 0 \Rightarrow t = \frac{21.65}{4.905} \approx 4.41 \, \text{s}
\]

4. **Range of the projectile:**
The horizontal range (\( R \)) can be determined using the time of flight:
\[
R = x(T) = v_{0x} \cdot T = 12.5 \cdot 4.41 \approx 55.125 \, \text{m}
\]

5. **Maximum height:**
The maximum height (\( H \)) occurs when the vertical velocity component is zero. The time to reach maximum height (\( t_{max} \)) can be calculated as:
\[
t_{max} = \frac{v_{0y}}{g} = \frac{21.65}{9.81} \approx 2.20 \, \text{s}
\]
Substitute \( t_{max} \) into the vertical motion equation to find \( H \):
\[
H = 21.65 \cdot 2.20 - 4.905 \cdot (2.20)^2
\]
\[
H \approx 21.65 \cdot 2.20 - 4.905 \cdot 4.84 \approx 47.63 - 23.73 \approx 23.9 \, \text{m}
\]

### Summary:
- The projectile reaches a maximum height of approximately **23.9 meters**.
- The total time of flight is approximately **4.41 seconds**.
- The horizontal range is approximately **55.13 meters**.

The trajectory follows a parabolic path described by the above equations.