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 projectile’s launch speed, we analyze its vertical motion. The highest point it reaches corresponds to the maximum height ((h_{\text{max}})), given as 13.5 m. At this point, the vertical velocity is zero.
Step 1: Use the Kinematic Equation
The kinematic equation for vertical motion is:
[
v_y^2 = v_{y0}^2 – 2 g h_{\text{max}}
]
At the peak, ( v_y = 0 ), so:
[
0 = v_{y0}^2 – 2 g h_{\text{max}}
]
Solving for ( v_{y0} ):
[
v_{y0} = \sqrt{2 g h_{\text{max}}}
]
Substituting ( g = 9.81 \, \text{m/s}^2 ) and ( h_{\text{max}} = 13.5 ) m:
[
v_{y0} = \sqrt{2 (9.81)(13.5)}
]
[
v_{y0} = \sqrt{264.87}
]
[
v_{y0} \approx 16.27 \, \text{m/s}
]
Step 2: Relate to Initial Speed
The initial vertical velocity component is:
[
v_{y0} = v_0 \sin \theta
]
Substituting ( \theta = 15.0^\circ ):
[
16.27 = v_0 \sin 15.0^\circ
]
Since ( \sin 15.0^\circ \approx 0.2588 ):
[
16.27 = v_0 (0.2588)
]
Solving for ( v_0 ):
[
v_0 = \frac{16.27}{0.2588}
]
[
v_0 \approx 62.9 \, \text{m/s}
]
Conclusion:
The projectile’s launch speed is 62.9 m/s.
Explanation:
The key to solving this problem is recognizing that the projectile’s highest point occurs when its vertical velocity becomes zero. By using kinematic equations, we determine the initial vertical velocity component required to reach 13.5 m. Then, using trigonometry, we solve for the total launch speed.