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 is :
To solve this problem, we will use the principles of projectile motion. Specifically, we will apply the equation for the maximum height of a projectile, which is influenced by its initial velocity, the angle of launch, and acceleration due to gravity.
Step 1: Maximum Height Equation
The formula for the maximum height $H_{\text{max}}$ of a projectile is:
$$
H_{\text{max}} = \frac{v_0^2 \sin^2(\theta)}{2g}
$$
Where:
- $v_0$ is the initial launch velocity (which we are trying to find),
- $\theta$ is the launch angle,
- $g$ is the acceleration due to gravity, which is approximately $9.8 \, \text{m/s}^2$,
- $H_{\text{max}}$ is the maximum height.
Step 2: Given Information
- The maximum height $H_{\text{max}} = 13.5 \, \text{m}$,
- The launch angle $\theta = 15.0^\circ$,
- $g = 9.8 \, \text{m/s}^2$.
We need to solve for the initial velocity $v_0$.
Step 3: Rearranging the Equation
Rearrange the formula to solve for $v_0$:
$$
v_0 = \sqrt{\frac{2gH_{\text{max}}}{\sin^2(\theta)}}
$$
Step 4: Plugging in the Values
Substitute the known values into the equation:
$$
v_0 = \sqrt{\frac{2(9.8)(13.5)}{\sin^2(15^\circ)}}
$$
First, calculate $\sin(15^\circ)$. Using a calculator:
$$
\sin(15^\circ) \approx 0.2588
$$
Now substitute this into the equation:
$$
v_0 = \sqrt{\frac{2(9.8)(13.5)}{(0.2588)^2}}
$$
$$
v_0 = \sqrt{\frac{264.6}{0.0670}} \approx \sqrt{3940.3}
$$
$$
v_0 \approx 62.8 \, \text{m/s}
$$
Final Answer:
The initial launch speed of the projectile is approximately 62.8 m/s.
Explanation:
In this problem, the key to finding the launch speed was understanding how the height of the projectile relates to its initial velocity. The formula used calculates the height based on the vertical component of the velocity, which is influenced by the sine of the launch angle. By rearranging the formula and inputting the given maximum height and launch angle, we derived the initial velocity needed to reach that height.