The range of a particle when launched at an angle of $15^\circ$ with the horizontal is $1.5 ext{ km}$. What i

Question

The range of a particle when launched at an angle of $15^\circ$ with the horizontal is $1.5	ext{ km}$. What is the range of the projectile when launched at an angle of $45^\circ$ to the horizontal?

Accepted Answer

1.  **Formula for Horizontal Range:** The horizontal range $R$ is given by $R = \frac{u^2 \sin 2	heta}{g}$, where $u$ is the projection speed and $	heta$ is the angle of projection .
2.  **Case 1 ($	heta = 15^\circ$):**
    $R_1 = \frac{u^2 \sin(2 	imes 15^\circ)}{g} = \frac{u^2 \sin 30^\circ}{g}$.
    Given $R_1 = 1.5	ext{ km}$ and $\sin 30^\circ = 0.5$, we have:
    $1.5 = \frac{u^2}{g} 	imes 0.5 \implies \frac{u^2}{g} = 3.0	ext{ km}$.
3.  **Case 2 ($	heta = 45^\circ$):**
    $R_2 = \frac{u^2 \sin(2 	imes 45^\circ)}{g} = \frac{u^2 \sin 90^\circ}{g}$.
    Since $\sin 90^\circ = 1$ and we found $\frac{u^2}{g} = 3.0	ext{ km}$:
    $R_2 = 3.0 	imes 1 = 3.0	ext{ km}$.

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