The highest barrier that a projectile can clear is 13.5 m, when the projectile is launched at an angle of 15.0 degrees above the horizontal. What is the projectiles launch speed?
The Correct Answer and Explanation
To determine the launch speed of a projectile that clears a barrier of 13.5 m at a launch angle of 15.0°, we need to analyze the vertical motion using kinematics.
Step 1: Use vertical motion formula
The formula for the maximum height $h$ of a projectile is:
$$
h = \frac{v_0^2 \sin^2\theta}{2g}
$$
Where:
- $h = 13.5 \, \text{m}$ (maximum height),
- $v_0$ = launch speed (unknown),
- $\theta = 15.0^\circ$,
- $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity),
- $\sin^2\theta = (\sin 15^\circ)^2$
Step 2: Plug values into the formula
$$
13.5 = \frac{v_0^2 (\sin 15^\circ)^2}{2 \cdot 9.8}
$$
$$
\sin 15^\circ \approx 0.2588 \quad \Rightarrow \quad (\sin 15^\circ)^2 \approx 0.067
$$
$$
13.5 = \frac{v_0^2 \cdot 0.067}{19.6}
$$
Multiply both sides by 19.6:
$$
13.5 \cdot 19.6 = v_0^2 \cdot 0.067
$$
$$
264.6 = v_0^2 \cdot 0.067
$$
Now divide both sides by 0.067:
$$
v_0^2 = \frac{264.6}{0.067} \approx 3,948.5
$$
$$
v_0 = \sqrt{3948.5} \approx \boxed{62.8 \, \text{m/s}}
$$
Explanation (300+ words)
To solve this problem, we focus on the vertical motion of the projectile. When a projectile is launched, its motion can be split into horizontal and vertical components. The maximum height that a projectile reaches is influenced only by its vertical velocity component and gravity.
The vertical component of the initial velocity is given by $v_0 \sin \theta$, and using basic kinematics, the formula for the maximum vertical height (assuming upward motion until vertical velocity becomes zero) is:
$$
h = \frac{(v_0 \sin \theta)^2}{2g}
$$
In this case, we know the projectile reaches a maximum height of 13.5 m, and it was launched at 15.0°, a relatively shallow angle. That means the vertical component of its velocity is relatively small, and it needs a large total launch speed to achieve that height.
By solving for $v_0$, we rearranged the formula and substituted known values: $g = 9.8 \, \text{m/s}^2$ and $\sin(15^\circ) \approx 0.2588$. Squaring the sine value, multiplying and isolating $v_0$, we find the launch speed to be approximately 62.8 m/s.
This result shows that even at a low angle, a high speed can allow the projectile to reach a significant height. Understanding how vertical and horizontal components affect projectile motion is key in solving such physics problems.