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 (( v_0 )), we use the kinematic equation for vertical motion:
[
v^2 = v_0^2 + 2 a s
]
where:
- ( v ) is the final vertical velocity at the highest point (( v = 0 ), since the projectile momentarily stops moving upward at the peak),
- ( v_0 ) is the initial vertical velocity component (( v_0 \sin \theta )),
- ( a = -9.81 \, \text{m/s}^2 ) (acceleration due to gravity),
- ( s = 13.5 ) m (maximum height).
Rewriting the equation:
[
0 = (v_0 \sin 15^\circ)^2 – 2(9.81)(13.5)
]
[
(v_0 \sin 15^\circ)^2 = 2(9.81)(13.5)
]
[
v_0^2 \sin^2 15^\circ = 264.87
]
[
v_0^2 = \frac{264.87}{\sin^2 15^\circ}
]
[
v_0^2 = \frac{264.87}{(0.2588)^2}
]
[
v_0^2 = \frac{264.87}{0.067}
]
[
v_0 = \sqrt{3954}
]
[
v_0 \approx 62.9 \text{ m/s}
]
Explanation:
A projectile follows a parabolic path due to gravity. At the highest point, its vertical velocity becomes zero. By analyzing only the vertical motion, we used the kinematic equation to solve for the initial speed. The vertical component of velocity depends on the angle of launch. Given the highest point of 13.5 m, we found the launch speed needed to reach that height. The correct answer is approximately 62.9 m/s.