Understanding Projectile Motion
In physics, a projectile is an object thrown into the air that moves under the influence of gravity. When we talk about projectile motion, we generally focus on two key metrics:
- Horizontal Range ($R$): The total horizontal distance covered by the object.
- Maximum Height ($H$): The highest vertical point reached by the object during its flight.
The Formulas
For a projectile launched with an initial velocity $v$ at an angle $\theta$ with the horizontal, the formulas are:
- Horizontal Range: $R = \frac{v^2 \sin(2\theta)}{g}$
- Maximum Height: $H = \frac{v^2 \sin^2\theta}{2g}$
(Where $g$ is the acceleration due to gravity, approx $9.8 \, m/s^2$)
Solving the Problem
We are asked to find the angle $\theta$ such that $R = H$. Setting the formulas equal to each other:
$$\frac{v^2 \sin(2\theta)}{g} = \frac{v^2 \sin^2\theta}{2g}$$
Step-by-Step Simplification:
Cancel common terms: Both sides have $v^2$ and $g$, so we can remove them. $$\sin(2\theta) = \frac{\sin^2\theta}{2}$$
Use the double-angle identity: Replace $\sin(2\theta)$ with $2\sin\theta\cos\theta$. $$2\sin\theta\cos\theta = \frac{\sin^2\theta}{2}$$
Rearrange to solve for $\tan\theta$: $$4\sin\theta\cos\theta = \sin^2\theta$$ Divide both sides by $\sin\theta\cos\theta$ (assuming $\theta \neq 0$): $$4 = \frac{\sin\theta}{\cos\theta}$$ $$\tan\theta = 4$$
Calculate the angle: $$\theta = \arctan(4) \approx 75.96^{\circ}$$
Conclusion
For a projectile to have equal horizontal range and maximum height, it must be launched at an angle of approximately $75.96^{\circ}$.