Sure! Let's break this problem down into steps:
### Given Data
- Initial velocity ([tex]\(v_0\)[/tex]): 150 m/s
- Angle ([tex]\(\theta\)[/tex]) with the horizontal: [tex]\(30^\circ\)[/tex]
- Height of the cliff ([tex]\(h_0\)[/tex]): 50 meters
- Acceleration due to gravity ([tex]\(g\)[/tex]): [tex]\(9.81 \, \text{m/s}^2\)[/tex]
### Conversion of Angle to Radians
First, let's convert the angle from degrees to radians since trigonometric functions in physics typically use radians.
[tex]\[
\theta_\text{rad} = \frac{30 \times \pi}{180} = \frac{\pi}{6}
\][/tex]
### Decomposition of Initial Velocity
The initial velocity is split into two components:
- Horizontal component ([tex]\(v_{0x}\)[/tex]):
[tex]\[
v_{0x} = v_0 \cos(\theta_\text{rad}) = 150 \cos\left(\frac{\pi}{6}\right) = 150 \times \frac{\sqrt{3}}{2} \approx 129.9 \, \text{m/s}
\][/tex]
- Vertical component ([tex]\(v_{0y}\)[/tex]):
[tex]\[
v_{0y} = v_0 \sin(\theta_\text{rad}) = 150 \sin\left(\frac{\pi}{6}\right) = 150 \times \frac{1}{2} = 75 \, \text{m/s}
\][/tex]
### Part (a): Time of Flight to the Ground
To find the time ([tex]\(t\)[/tex]) at which the projectile hits the ground, we use the vertical motion equation:
[tex]\[
h = h_0 + v_{0y} t - \frac{1}{2} g t^2
\][/tex]
Given that the final height [tex]\(h = 0\)[/tex] (ground level), we rewrite it as a quadratic equation:
[tex]\[
0 = 50 + 75t - \frac{1}{2} \cdot 9.81 t^2
\][/tex]
This simplifies to:
[tex]\[
4.905 t^2 - 75t - 50 = 0
\][/tex]
To solve this quadratic equation, we use the quadratic formula:
[tex]\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Where [tex]\(a = -4.905\)[/tex], [tex]\(b = 75\)[/tex], [tex]\(c = 50\)[/tex]:
[tex]\[
t = \frac{-75 \pm \sqrt{75^2 - 4(-4.905)(50)}}{2(-4.905)}
\][/tex]
Calculating the discriminant:
[tex]\[
75^2 - 4 \cdot (-4.905) \cdot 50 = 5625 + 981 = 6606
\][/tex]
Thus:
[tex]\[
t = \frac{-75 \pm \sqrt{6606}}{-9.81}
\][/tex]
Solving for the two values:
[tex]\[
t_1 = \frac{-75 + 81.26}{-9.81} \approx -0.64 \, \text{s} \quad \text{(not meaningful as time cannot be negative)}
\][/tex]
[tex]\[
t_2 = \frac{-75 - 81.26}{-9.81} \approx 15.93 \, \text{s}
\][/tex]
The projectile takes approximately [tex]\(15.93\)[/tex] seconds to hit the ground.
### Horizontal Distance
To find the horizontal distance traveled ([tex]\(d\)[/tex]):
[tex]\[
d = v_{0x} \times t = 129.9 \, \text{m/s} \times 15.93 \, \text{s} \approx 2069.42 \, \text{m}
\][/tex]
### Part (b): Greatest Elevation Above the Ground
The greatest elevation occurs at the peak of the trajectory.
The time to reach maximum height ([tex]\(t_\text{max}\)[/tex]) is when the vertical velocity component becomes zero:
[tex]\[
t_\text{max} = \frac{v_{0y}}{g} = \frac{75}{9.81} \approx 7.65 \, \text{s}
\][/tex]
Max height ([tex]\(H\)[/tex]) above the cliff:
[tex]\[
H = h_0 + v_{0y} t_\text{max} - \frac{1}{2} g t_\text{max}^2
\][/tex]
[tex]\[
H = 50 + 75 \cdot 7.65 - \frac{1}{2} \cdot 9.81 \cdot 7.65^2
\][/tex]
[tex]\[
H = 50 + 573.75 - 286.05 \approx 336.7 \, \text{m}
\][/tex]
Therefore, the greatest elevation above the ground is approximately [tex]\(336.7\)[/tex] meters.
