The horizontal range of a projectile is maximum when the angle of projection is ____.
Show Hint
At 45°, the projectile has the perfect balance between vertical height (to stay in the air longer) and horizontal velocity (to move forward faster). At angles like 90°, you get height but no forward distance!
Step 1: Understanding the Concept
The horizontal range is the total distance covered by a projectile along the horizontal plane. It depends on the initial velocity and the angle at which the object is launched relative to the ground. Step 2: Key Formula or Approach
The formula for horizontal range ($R$) is:
\[ R = \frac{u^2 \sin(2\theta)}{g} \] Step 3: Detailed Explanation
1. To maximize $R$, the term $\sin(2\theta)$ must be at its maximum possible value.
2. The maximum value for any sine function is 1.
3. We set $\sin(2\theta) = 1$. Since $\sin(90^\circ) = 1$, we have:
\[ 2\theta = 90^\circ \]
4. Solving for $\theta$: $\theta = 45^\circ$. Step 4: Final Answer
The horizontal range is maximum when the angle of projection is 45°.