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 clears a barrier of 13.5 meters when launched at an angle of 15.0 degrees above the horizontal, we can apply the principles of projectile motion. The key concept here is the relationship between the maximum height reached by the projectile and its initial velocity components.
Step 1: Understand the Maximum Height Formula
The maximum height ( h ) reached by a projectile is given by the formula:
[
h = \frac{v_0^2 \sin^2 \theta}{2g}
]
Where:
- ( v_0 ) is the initial launch speed,
- ( \theta ) is the launch angle,
- ( g ) is the acceleration due to gravity (approximately 9.81 m/s²).
Step 2: Rearrange the Formula to Solve for ( v_0 )
Rearranging the above formula to solve for ( v_0 ), we get:
[
v_0 = \sqrt{\frac{2gh}{\sin^2 \theta}}
]
Step 3: Substitute Known Values
Given:
- Maximum height ( h = 13.5 ) meters,
- Launch angle ( \theta = 15.0^\circ ),
- Acceleration due to gravity ( g = 9.81 ) m/s².
First, calculate ( \sin(15^\circ) ):
[
\sin(15^\circ) \approx 0.2588
]
Now, substitute these values into the equation for ( v_0 ):
[
v_0 = \sqrt{\frac{2 \times 9.81 \times 13.5}{(0.2588)^2}}
]
Step 4: Perform the Calculations
Calculate the numerator:
[
2 \times 9.81 \times 13.5 = 264.87
]
Calculate the denominator:
[
(0.2588)^2 \approx 0.0670
]
Now, compute ( v_0 ):
[
v_0 = \sqrt{\frac{264.87}{0.0670}} \approx \sqrt{3953.73} \approx 62.88 \, \text{m/s}
]
Conclusion
The initial launch speed required for the projectile to clear a barrier of 13.5 meters at a launch angle of 15.0 degrees is approximately 62.88 m/s. This calculation aligns with the solution provided by Vaia’s textbook. citeturn0search0
Understanding this relationship is crucial in projectile motion problems, as it allows us to determine the necessary launch parameters to achieve a desired trajectory or clearance height.