Question
A projectile is Fired horizontally from the top of a cliff 250m high. Projectile lands 1414m from the bottom of the cliff. Find:
The inital speed of the projectile
The velocity of the projectile as it hits the ground
The inital speed of the projectile
The velocity of the projectile as it hits the ground
Answers
GPT-4o mini
To solve this problem, we need to analyze the projectile's motion separately in the horizontal (x) and vertical (y) directions.
### 1. Vertical Motion
The projectile is fired horizontally, so its initial vertical velocity (\(v_{0y}\)) is \(0 \, \text{m/s}\). The only force acting on it is gravity. We can use the following kinematic equation for the vertical motion:
\[
y = v_{0y} t + \frac{1}{2} g t^2
\]
where:
- \(y = 250 \, \text{m}\) (the height of the cliff, downward is positive)
- \(g = 9.81 \, \text{m/s}^2\) (acceleration due to gravity)
- \(t\) is the time of flight.
Substituting the known values, we get:
\[
250 = 0 \cdot t + \frac{1}{2} (9.81) t^2
\]
Simplifying this gives:
\[
250 = \frac{1}{2} (9.81) t^2
\]
\[
250 = 4.905 t^2
\]
\[
t^2 = \frac{250}{4.905}
\]
\[
t^2 \approx 50.98
\]
\[
t \approx \sqrt{50.98} \approx 7.14 \, \text{s}
\]
### 2. Horizontal Motion
Next, we can determine the initial horizontal speed (\(v_{0x}\)) by using the horizontal distance the projectile traveled, which is \(1414 \, \text{m}\):
The horizontal motion can be described by the formula:
\[
x = v_{0x} t
\]
We know:
- \(x = 1414 \, \text{m}\)
- \(t \approx 7.14 \, \text{s}\)
Substituting the known values:
\[
1414 = v_{0x} (7.14)
\]
Solving for \(v_{0x}\):
\[
v_{0x} = \frac{1414}{7.14} \approx 198.7 \, \text{m/s}
\]
### 3. Velocity of the Projectile as it Hits the Ground
Finally, we need to calculate the velocity of the projectile just before it hits the ground, which can be found by combining the horizontal and vertical components of the velocity at that time.
- The horizontal component (\(v_{x}\)) remains constant and is equal to \(v_{0x}\):
\[
v_{x} = 198.7 \, \text{m/s}
\]
- The vertical component (\(v_{y}\)) can be found using the formula:
\[
v_{y} = v_{0y} + gt
\]
Since \(v_{0y} = 0\):
\[
v_{y} = 0 + (9.81)(7.14) \approx 70.0 \, \text{m/s}
\]
### 4. Resultant Velocity
To find the resultant velocity (\(v\)) at which the projectile hits the ground, we can use Pythagoras' theorem:
\[
v = \sqrt{v_{x}^2 + v_{y}^2}
\]
Calculating this gives:
\[
v = \sqrt{(198.7)^2 + (70.0)^2}
\]
Calculating each part:
\[
v = \sqrt{39486.69 + 4900} \approx \sqrt{44386.69} \approx 210.75 \, \text{m/s}
\]
### Summary of Results
- The initial speed of the projectile (\(v_{0x}\)) is approximately **198.7 m/s**.
- The velocity of the projectile as it hits the ground is approximately **210.75 m/s**.
### 1. Vertical Motion
The projectile is fired horizontally, so its initial vertical velocity (\(v_{0y}\)) is \(0 \, \text{m/s}\). The only force acting on it is gravity. We can use the following kinematic equation for the vertical motion:
\[
y = v_{0y} t + \frac{1}{2} g t^2
\]
where:
- \(y = 250 \, \text{m}\) (the height of the cliff, downward is positive)
- \(g = 9.81 \, \text{m/s}^2\) (acceleration due to gravity)
- \(t\) is the time of flight.
Substituting the known values, we get:
\[
250 = 0 \cdot t + \frac{1}{2} (9.81) t^2
\]
Simplifying this gives:
\[
250 = \frac{1}{2} (9.81) t^2
\]
\[
250 = 4.905 t^2
\]
\[
t^2 = \frac{250}{4.905}
\]
\[
t^2 \approx 50.98
\]
\[
t \approx \sqrt{50.98} \approx 7.14 \, \text{s}
\]
### 2. Horizontal Motion
Next, we can determine the initial horizontal speed (\(v_{0x}\)) by using the horizontal distance the projectile traveled, which is \(1414 \, \text{m}\):
The horizontal motion can be described by the formula:
\[
x = v_{0x} t
\]
We know:
- \(x = 1414 \, \text{m}\)
- \(t \approx 7.14 \, \text{s}\)
Substituting the known values:
\[
1414 = v_{0x} (7.14)
\]
Solving for \(v_{0x}\):
\[
v_{0x} = \frac{1414}{7.14} \approx 198.7 \, \text{m/s}
\]
### 3. Velocity of the Projectile as it Hits the Ground
Finally, we need to calculate the velocity of the projectile just before it hits the ground, which can be found by combining the horizontal and vertical components of the velocity at that time.
- The horizontal component (\(v_{x}\)) remains constant and is equal to \(v_{0x}\):
\[
v_{x} = 198.7 \, \text{m/s}
\]
- The vertical component (\(v_{y}\)) can be found using the formula:
\[
v_{y} = v_{0y} + gt
\]
Since \(v_{0y} = 0\):
\[
v_{y} = 0 + (9.81)(7.14) \approx 70.0 \, \text{m/s}
\]
### 4. Resultant Velocity
To find the resultant velocity (\(v\)) at which the projectile hits the ground, we can use Pythagoras' theorem:
\[
v = \sqrt{v_{x}^2 + v_{y}^2}
\]
Calculating this gives:
\[
v = \sqrt{(198.7)^2 + (70.0)^2}
\]
Calculating each part:
\[
v = \sqrt{39486.69 + 4900} \approx \sqrt{44386.69} \approx 210.75 \, \text{m/s}
\]
### Summary of Results
- The initial speed of the projectile (\(v_{0x}\)) is approximately **198.7 m/s**.
- The velocity of the projectile as it hits the ground is approximately **210.75 m/s**.