Question

projectile motion problem clears barrier 38° angle initial velocity

Answer 1

To analyze a projectile motion problem where a projectile is launched at an angle of 38° and needs to clear a barrier, we typically consider the following steps:

### Given Parameters
1. Launch angle, \(\theta = 38^\circ\)
2. Initial velocity, \(v_0\) (this value must be specified or calculated)
3. Height of the barrier, \(h\) (also needs to be given)
4. Horizontal distance to the barrier, \(d\)

### Equations of Motion
The motion can be broken down into horizontal and vertical components:

1. **Horizontal motion**:
- The horizontal component of the initial velocity:
\[
v_{0x} = v_0 \cdot \cos(\theta)
\]
- Horizontal displacement after time \(t\):
\[
x(t) = v_{0x} \cdot t
\]

2. **Vertical motion**:
- The vertical component of the initial velocity:
\[
v_{0y} = v_0 \cdot \sin(\theta)
\]
- Vertical displacement after time \(t\):
\[
y(t) = v_{0y} \cdot t - \frac{1}{2} g t^2
\]
where \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)).

### Reaching the Barrier
To determine if the projectile clears the barrier, we need to find the time \(t\) when the projectile reaches the horizontal position \(d\):

1. Set \(x(t) = d\):
\[
d = v_{0x} \cdot t \implies t = \frac{d}{v_{0} \cdot \cos(\theta)}
\]

2. Substitute \(t\) into the vertical motion equation to find the height \(y\) at which the projectile is at the barrier:
\[
y\left(\frac{d}{v_{0} \cos(\theta)}\right) = v_{0y} \cdot \left(\frac{d}{v_{0} \cos(\theta)}\right) - \frac{1}{2} g \left(\frac{d}{v_{0} \cos(\theta)}\right)^2
\]

3. Simplifying this gives:
\[
y(d) = (v_0 \sin(\theta)) \left(\frac{d}{v_{0} \cos(\theta)}\right) - \frac{1}{2} g \left(\frac{d}{v_{0} \cos(\theta)}\right)^2
\]
\[
= \frac{d \cdot \tan(\theta)}{1} - \frac{g d^2}{2 v_{0}^2 \cos^2(\theta)}
\]
\[
= d \tan(38^\circ) - \frac{g d^2}{2 v_{0}^2 \cos^2(38^\circ)}
\]

### Condition to Clear the Barrier
To clear the barrier, the height \(y(d)\) must be greater than or equal to the height of the barrier \(h\):
\[
d \tan(38^\circ) - \frac{g d^2}{2 v_{0}^2 \cos^2(38^\circ)} \geq h
\]

From here, you can solve for the initial velocity \(v_0\) necessary to clear the barrier or check if a given initial velocity is sufficient. If you have the specific height \(h\) of the barrier and the horizontal distance \(d\) to the barrier, you can proceed with these calculations. If you provide those values, I can help with further calculations!