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 find the launch speed of a projectile that reaches a maximum height of 13.5 meters when launched at an angle of 15.0° above the horizontal, we can use kinematic equations from projectile motion. Let’s start with the solution and then give a full explanation.
✅ Correct Answer:
Launch speed = 34.2 m/s
🧠 Explanation (300+ words):
Projectile motion is a type of two-dimensional motion under constant acceleration due to gravity. It can be analyzed by breaking it into horizontal and vertical components. In this question, we are given the maximum height of a projectile and the angle of launch, and we are asked to determine the initial speed at which it was launched.
Step 1: Analyze vertical motion
At the maximum height of the projectile, the vertical velocity component (Vy) becomes zero. That means all vertical kinetic energy has converted to gravitational potential energy.
We use the kinematic equation for vertical motion:
$$
v_y^2 = v_{y0}^2 – 2g h
$$
At the highest point, $v_y = 0$, so:
$$
0 = v_{y0}^2 – 2 g h
\Rightarrow v_{y0}^2 = 2 g h
\Rightarrow v_{y0} = \sqrt{2gh}
$$
Where:
- $v_{y0} = v_0 \sin(\theta)$ is the initial vertical velocity
- $g = 9.8 \, \text{m/s}^2$
- $h = 13.5 \, \text{m}$
- $\theta = 15^\circ$
First, solve for $v_{y0}$:
$$
v_{y0} = \sqrt{2 \cdot 9.8 \cdot 13.5} = \sqrt{264.6} \approx 16.27 \, \text{m/s}
$$
Now, relate this to the full initial velocity $v_0$:
$$
v_0 \sin(15^\circ) = 16.27
$$
$$
v_0 = \frac{16.27}{\sin(15^\circ)} = \frac{16.27}{0.2588} \approx 62.86 \, \text{m/s}
$$
⚠️ But wait — we notice something is off. Let’s re-check. The vertical component was:
$$
v_{y0} = v_0 \sin(15^\circ)
$$
So:
$$
v_0 = \frac{v_{y0}}{\sin(15^\circ)} = \frac{16.27}{0.2588} \approx \boxed{62.86 \, \text{m/s}}
$$
✔️ Final corrected launch speed is approximately 62.9 m/s, not 34.2 m/s as initially stated.
✅ Final Answer:
Launch speed ≈ 62.9