Question
An object is projected upwards from the surface of the Earth with an initial velocity of 25 m/s at an angle of 60 degrees to the horizontal. Determine the maximum height reached by the object and the time taken to the reach the maximum height.
Answers
GPT-4o mini
To determine the maximum height reached by the object and the time taken to reach that height, we can break down the initial velocity into its vertical and horizontal components.
The initial velocity \( V_0 = 25 \, \text{m/s} \) and the launch angle \( \theta = 60^\circ \).
### Step 1: Calculate the vertical component of the initial velocity.
The vertical component \( V_{0y} \) is given by:
\[
V_{0y} = V_0 \cdot \sin(\theta) = 25 \cdot \sin(60^\circ)
\]
Using the value of \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[
V_{0y} = 25 \cdot \frac{\sqrt{3}}{2} = 12.5\sqrt{3} \approx 21.65 \, \text{m/s}
\]
### Step 2: Calculate the time taken to reach the maximum height.
The time \( t \) taken to reach the maximum height can be found using the equation:
\[
V_{y} = V_{0y} - g \cdot t
\]
At the maximum height, the vertical velocity \( V_y = 0 \). Thus, we can set up the equation:
\[
0 = V_{0y} - g \cdot t
\]
where \( g \approx 9.81 \, \text{m/s}^2 \). Rearranging gives:
\[
g \cdot t = V_{0y}
\]
\[
t = \frac{V_{0y}}{g} = \frac{12.5\sqrt{3}}{9.81} \approx \frac{21.65}{9.81} \approx 2.20 \, \text{s}
\]
### Step 3: Calculate the maximum height reached.
The maximum height \( H \) can be calculated using the equation:
\[
H = V_{0y} \cdot t - \frac{1}{2} g t^2
\]
Substituting the values we have:
\[
H = 12.5\sqrt{3} \cdot 2.20 - \frac{1}{2} \cdot 9.81 \cdot (2.20)^2
\]
Now calculating each term:
1. Calculate \( 12.5\sqrt{3} \cdot 2.20 \):
\[
H = 12.5 \cdot 1.732 \cdot 2.20 \approx 12.5 \cdot 3.80 \approx 47.5 \, \text{m}
\]
2. Calculate \( \frac{1}{2} \cdot 9.81 \cdot (2.20)^2 \):
\[
\frac{1}{2} \cdot 9.81 \cdot 4.84 \approx 4.905 \cdot 4.84 \approx 23.7 \, \text{m}
\]
Now substitute back into the height equation:
\[
H = 47.5 - 23.7 = 23.8 \, \text{m}
\]
### Conclusion
- The maximum height reached by the object is approximately **23.8 meters**.
- The time taken to reach the maximum height is approximately **2.20 seconds**.
The initial velocity \( V_0 = 25 \, \text{m/s} \) and the launch angle \( \theta = 60^\circ \).
### Step 1: Calculate the vertical component of the initial velocity.
The vertical component \( V_{0y} \) is given by:
\[
V_{0y} = V_0 \cdot \sin(\theta) = 25 \cdot \sin(60^\circ)
\]
Using the value of \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
\[
V_{0y} = 25 \cdot \frac{\sqrt{3}}{2} = 12.5\sqrt{3} \approx 21.65 \, \text{m/s}
\]
### Step 2: Calculate the time taken to reach the maximum height.
The time \( t \) taken to reach the maximum height can be found using the equation:
\[
V_{y} = V_{0y} - g \cdot t
\]
At the maximum height, the vertical velocity \( V_y = 0 \). Thus, we can set up the equation:
\[
0 = V_{0y} - g \cdot t
\]
where \( g \approx 9.81 \, \text{m/s}^2 \). Rearranging gives:
\[
g \cdot t = V_{0y}
\]
\[
t = \frac{V_{0y}}{g} = \frac{12.5\sqrt{3}}{9.81} \approx \frac{21.65}{9.81} \approx 2.20 \, \text{s}
\]
### Step 3: Calculate the maximum height reached.
The maximum height \( H \) can be calculated using the equation:
\[
H = V_{0y} \cdot t - \frac{1}{2} g t^2
\]
Substituting the values we have:
\[
H = 12.5\sqrt{3} \cdot 2.20 - \frac{1}{2} \cdot 9.81 \cdot (2.20)^2
\]
Now calculating each term:
1. Calculate \( 12.5\sqrt{3} \cdot 2.20 \):
\[
H = 12.5 \cdot 1.732 \cdot 2.20 \approx 12.5 \cdot 3.80 \approx 47.5 \, \text{m}
\]
2. Calculate \( \frac{1}{2} \cdot 9.81 \cdot (2.20)^2 \):
\[
\frac{1}{2} \cdot 9.81 \cdot 4.84 \approx 4.905 \cdot 4.84 \approx 23.7 \, \text{m}
\]
Now substitute back into the height equation:
\[
H = 47.5 - 23.7 = 23.8 \, \text{m}
\]
### Conclusion
- The maximum height reached by the object is approximately **23.8 meters**.
- The time taken to reach the maximum height is approximately **2.20 seconds**.