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 determine the launch speed of a projectile that reaches a maximum height of 13.5 m when launched at an angle of 15.0°, we need to apply principles of projectile motion and kinematics.
🔹 Final Answer:
The projectile’s launch speed is approximately 37.9 m/s.
🔹 Step-by-Step Explanation (300+ words):
A projectile reaches its maximum height when its vertical velocity component becomes zero (i.e., at the peak of the trajectory). We can use kinematic equations to analyze this vertical motion.
Let’s define the known variables:
- Maximum height, ( h = 13.5 \, \text{m} )
- Launch angle, ( \theta = 15.0^\circ )
- At the highest point, vertical velocity ( v_y = 0 \, \text{m/s} )
- Acceleration due to gravity, ( g = 9.8 \, \text{m/s}^2 )
Let ( v_0 ) be the initial launch speed, and ( v_{0y} ) be the initial vertical velocity component, which is given by:
[
v_{0y} = v_0 \cdot \sin(\theta)
]
We use the following kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:
[
v_y^2 = v_{0y}^2 – 2g h
]
Since ( v_y = 0 ) at the peak:
[
0 = v_{0y}^2 – 2gh
\Rightarrow v_{0y}^2 = 2gh
\Rightarrow v_{0y} = \sqrt{2gh}
]
Substitute the known values:
[
v_{0y} = \sqrt{2 \cdot 9.8 \cdot 13.5}
= \sqrt{264.6}
\approx 16.27 \, \text{m/s}
]
Now recall that:
[
v_{0y} = v_0 \cdot \sin(15^\circ)
\Rightarrow v_0 = \frac{v_{0y}}{\sin(15^\circ)}
]
Using ( \sin(15^\circ) \approx 0.2588 ):
[
v_0 = \frac{16.27}{0.2588} \approx 62.9 \, \text{m/s}
]
🔁 Correction: There was an error in approximating earlier. Let’s redo the calculation:
[
v_{0y} = \sqrt{2 \cdot 9.8 \cdot 13.5} = \sqrt{264.6} \approx 16.27 \, \text{m/s}
]
[
v_0 = \frac{16.27}{\sin(15^\circ)} = \frac{16.27}{0.2588} \approx \boxed{62.9 \, \text{m/s}}
]