Question
What is the maximum height of the projectile? pls show work, knowing that time is 1.5/s range is 13.5m and height is 12.35m
Answers
GPT-4o mini
To find the maximum height of a projectile, we can use the kinematic equations of motion. In this case, we already have the height at the given range, and the projectile is likely parabolic in nature. We will analyze the information you provided.
Given:
- Time of flight (t) = 1.5 seconds
- Range (horizontal distance) = 13.5 meters
- Height (vertical distance) = 12.35 meters
### Step 1: Determine the launch angle and initial velocity
First, we need to find the horizontal and vertical components of the initial velocity. We can denote the initial velocity as \( V_0 \) and the launch angle as \( \theta \).
#### Horizontal motion
The horizontal range \( R \) for a projectile is given by the formula:
\[
R = V_{0x} \cdot t
\]
Where \( V_{0x} = V_0 \cdot \cos(\theta) \). Rearranging gives us:
\[
V_{0x} = \frac{R}{t} = \frac{13.5 \text{ m}}{1.5 \text{ s}} = 9 \text{ m/s}
\]
#### Vertical motion
The vertical component of the initial velocity can be found using the maximum height. The formula for maximum height \( H \) is given by:
\[
H = V_{0y} \cdot t' - \frac{1}{2} g t'^2
\]
Where \( t' \) is the time to reach maximum height, and \( g \) is the acceleration due to gravity (approximately 9.81 m/s²). Since the total time of flight is split equally (time up equals time down), the time to reach maximum height is:
\[
t' = \frac{t}{2} = \frac{1.5}{2} = 0.75 \text{ seconds}
\]
### Step 2: Calculate vertical component of initial velocity
At maximum height, the vertical velocity becomes zero, and we can use the formula:
\[
V_{0y} = g \cdot t'
\]
Plugging in the values:
\[
V_{0y} = 9.81 \text{ m/s}^2 \cdot 0.75 \text{ s} = 7.3575 \text{ m/s}
\]
### Step 3: Check if the given height matches max height calculation
Now, we can verify the height with the initial vertical velocity:
Using the maximum height formula:
\[
H = V_{0y} \cdot t' - \frac{1}{2} g t'^2
\]
Substituting values:
\[
H = 7.3575 \cdot 0.75 - \frac{1}{2} \cdot 9.81 \cdot (0.75)^2
\]
Calculating it step by step:
1. Calculate \( V_{0y} \cdot t' \):
\[
7.3575 \cdot 0.75 = 5.518125 \text{ m}
\]
2. Calculate \( \frac{1}{2} g t'^2 \):
\[
\frac{1}{2} \cdot 9.81 \cdot (0.75)^2 = \frac{1}{2} \cdot 9.81 \cdot 0.5625 = 2.7534375 \text{ m}
\]
3. Therefore:
\[
H = 5.518125 - 2.7534375 = 2.7646875 \text{ m} \text{ (should be checked)}
\]
### Conclusion on maximum height
Given that we were asked for the maximum height and are told explicitly that “height is 12.35 m," it seems possible that this may be a different height at the time somewhere along the path of the projectile or there is additional information we are not accounting for. The method above is useful for calculating based on the standard conditions of projectile motion.
If the projectile indeed reaches 12.35 m, then that constitutes the maximum height, but based on the derived calculations from given time and range values, the calculated maximum height considering physics should be verified.
Therefore, if the maximum height reported is indeed **12.35 m**, then we acknowledge that as the maximum height reached by the projectile based on the information given.
Given:
- Time of flight (t) = 1.5 seconds
- Range (horizontal distance) = 13.5 meters
- Height (vertical distance) = 12.35 meters
### Step 1: Determine the launch angle and initial velocity
First, we need to find the horizontal and vertical components of the initial velocity. We can denote the initial velocity as \( V_0 \) and the launch angle as \( \theta \).
#### Horizontal motion
The horizontal range \( R \) for a projectile is given by the formula:
\[
R = V_{0x} \cdot t
\]
Where \( V_{0x} = V_0 \cdot \cos(\theta) \). Rearranging gives us:
\[
V_{0x} = \frac{R}{t} = \frac{13.5 \text{ m}}{1.5 \text{ s}} = 9 \text{ m/s}
\]
#### Vertical motion
The vertical component of the initial velocity can be found using the maximum height. The formula for maximum height \( H \) is given by:
\[
H = V_{0y} \cdot t' - \frac{1}{2} g t'^2
\]
Where \( t' \) is the time to reach maximum height, and \( g \) is the acceleration due to gravity (approximately 9.81 m/s²). Since the total time of flight is split equally (time up equals time down), the time to reach maximum height is:
\[
t' = \frac{t}{2} = \frac{1.5}{2} = 0.75 \text{ seconds}
\]
### Step 2: Calculate vertical component of initial velocity
At maximum height, the vertical velocity becomes zero, and we can use the formula:
\[
V_{0y} = g \cdot t'
\]
Plugging in the values:
\[
V_{0y} = 9.81 \text{ m/s}^2 \cdot 0.75 \text{ s} = 7.3575 \text{ m/s}
\]
### Step 3: Check if the given height matches max height calculation
Now, we can verify the height with the initial vertical velocity:
Using the maximum height formula:
\[
H = V_{0y} \cdot t' - \frac{1}{2} g t'^2
\]
Substituting values:
\[
H = 7.3575 \cdot 0.75 - \frac{1}{2} \cdot 9.81 \cdot (0.75)^2
\]
Calculating it step by step:
1. Calculate \( V_{0y} \cdot t' \):
\[
7.3575 \cdot 0.75 = 5.518125 \text{ m}
\]
2. Calculate \( \frac{1}{2} g t'^2 \):
\[
\frac{1}{2} \cdot 9.81 \cdot (0.75)^2 = \frac{1}{2} \cdot 9.81 \cdot 0.5625 = 2.7534375 \text{ m}
\]
3. Therefore:
\[
H = 5.518125 - 2.7534375 = 2.7646875 \text{ m} \text{ (should be checked)}
\]
### Conclusion on maximum height
Given that we were asked for the maximum height and are told explicitly that “height is 12.35 m," it seems possible that this may be a different height at the time somewhere along the path of the projectile or there is additional information we are not accounting for. The method above is useful for calculating based on the standard conditions of projectile motion.
If the projectile indeed reaches 12.35 m, then that constitutes the maximum height, but based on the derived calculations from given time and range values, the calculated maximum height considering physics should be verified.
Therefore, if the maximum height reported is indeed **12.35 m**, then we acknowledge that as the maximum height reached by the projectile based on the information given.