Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

1.5 km

3.0 km

6.0 km

0.75 km

Step-by-Step Solution

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

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