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 clears a maximum height (vertical displacement) of 13.5 m when launched at an angle of 15.0° above the horizontal, we can use kinematics.
Step 1: Use vertical motion to find vertical velocity
The vertical component of the initial velocity can be related to the maximum height using the following kinematic equation:
$$
v_y^2 = v_{y0}^2 – 2g h
$$
At the highest point, the vertical velocity $v_y$ = 0 (because the projectile momentarily stops moving upward), so:
$$
0 = v_{y0}^2 – 2gh
\Rightarrow v_{y0}^2 = 2gh
\Rightarrow v_{y0} = \sqrt{2gh}
$$
Plug in:
- $g = 9.8 \, \text{m/s}^2$
- $h = 13.5 \, \text{m}$
$$
v_{y0} = \sqrt{2 \cdot 9.8 \cdot 13.5} = \sqrt{264.6} \approx 16.27 \, \text{m/s}
$$
Step 2: Use angle to find total launch speed
Now we use the angle to relate vertical velocity to total velocity.
$$
v_{y0} = v_0 \sin \theta
\Rightarrow v_0 = \frac{v_{y0}}{\sin \theta}
$$
$$
\theta = 15^\circ \Rightarrow \sin 15^\circ \approx 0.2588
$$
$$
v_0 = \frac{16.27}{0.2588} \approx 62.87 \, \text{m/s}
$$
✅ Final Answer:
$$
\boxed{62.9 \, \text{m/s}}
$$
Explanation (300+ words)
In projectile motion, the maximum height a projectile reaches is determined solely by the vertical component of its velocity, since only vertical motion is influenced by gravity. At the top of its arc, the projectile’s vertical velocity becomes zero — it stops rising and begins to fall back down. This is a key point in solving this problem.
We’re told the projectile reaches a maximum vertical height of 13.5 m, and it was launched at an angle of 15°. Since only the vertical part of the launch speed contributes to reaching this height, we start by using the kinematic equation that relates initial vertical velocity ($v_{y0}$) to the height. The equation $v_y^2 = v_{y0}^2 – 2gh$ simplifies to $v_{y0}^2 = 2gh$ when $v_y = 0$ at the peak.
After calculating the vertical component of the initial velocity to be about 16.27 m/s, we then use trigonometry to relate this vertical component to the full launch velocity. The sine of the launch angle gives us this relationship: $v_{y0} = v_0 \sin \theta$. By isolating $v_0$, we can solve for the full initial speed.
Using $\sin(15°) \approx 0.2588$, the total launch speed comes out to approximately 62.9 m/s. This is the speed required to launch the projectile at a 15° angle and still clear a vertical height of 13.5 m.
This problem is a classic example of how kinematics and trigonometry combine in physics to analyze motion in two dimensions.
