Question
An object is launched from the ground with an initial velocity of 50 m/s at an angle of 60 degrees. What is the horizontal distance covered by the object?
Answers
GPT-4o mini
To find the horizontal distance covered by the object, we first need to determine the time of flight and the horizontal component of the velocity.
1. **Initial Velocity Components**:
- The initial velocity (v₀) is given as 50 m/s.
- The angle of launch (θ) is 60 degrees.
The horizontal (v₀x) and vertical (v₀y) components can be calculated as follows:
\[
v_{0x} = v_0 \cdot \cos(\theta) = 50 \cdot \cos(60^\circ) = 50 \cdot 0.5 = 25 \ \text{m/s}
\]
\[
v_{0y} = v_0 \cdot \sin(\theta) = 50 \cdot \sin(60^\circ) = 50 \cdot \frac{\sqrt{3}}{2} \approx 43.3 \ \text{m/s}
\]
2. **Time of Flight (T)**:
The time of flight can be found using the vertical motion. The formula for time of flight when the object returns to the same vertical level is given by:
\[
T = \frac{2 v_{0y}}{g}
\]
where \(g\) is the acceleration due to gravity (approximately \(9.81 \ \text{m/s}^2\)).
\[
T = \frac{2 \cdot 43.3}{9.81} \approx \frac{86.6}{9.81} \approx 8.82 \ \text{s}
\]
3. **Horizontal Distance (Range)**:
The horizontal distance (range, R) covered by the object can be calculated by:
\[
R = v_{0x} \cdot T
\]
\[
R = 25 \ \text{m/s} \cdot 8.82 \ \text{s} \approx 220.5 \ \text{m}
\]
Thus, the horizontal distance covered by the object is approximately **220.5 meters**.
1. **Initial Velocity Components**:
- The initial velocity (v₀) is given as 50 m/s.
- The angle of launch (θ) is 60 degrees.
The horizontal (v₀x) and vertical (v₀y) components can be calculated as follows:
\[
v_{0x} = v_0 \cdot \cos(\theta) = 50 \cdot \cos(60^\circ) = 50 \cdot 0.5 = 25 \ \text{m/s}
\]
\[
v_{0y} = v_0 \cdot \sin(\theta) = 50 \cdot \sin(60^\circ) = 50 \cdot \frac{\sqrt{3}}{2} \approx 43.3 \ \text{m/s}
\]
2. **Time of Flight (T)**:
The time of flight can be found using the vertical motion. The formula for time of flight when the object returns to the same vertical level is given by:
\[
T = \frac{2 v_{0y}}{g}
\]
where \(g\) is the acceleration due to gravity (approximately \(9.81 \ \text{m/s}^2\)).
\[
T = \frac{2 \cdot 43.3}{9.81} \approx \frac{86.6}{9.81} \approx 8.82 \ \text{s}
\]
3. **Horizontal Distance (Range)**:
The horizontal distance (range, R) covered by the object can be calculated by:
\[
R = v_{0x} \cdot T
\]
\[
R = 25 \ \text{m/s} \cdot 8.82 \ \text{s} \approx 220.5 \ \text{m}
\]
Thus, the horizontal distance covered by the object is approximately **220.5 meters**.