The range of the projectile projected at an angle of 15$^\circ$ with horizontal is 50 m. If the projectile is projected with same velocity at an angle of 45$^\circ$ with horizontal, then its range will be

<p>The range $R$ of a projectile launched with an initial speed $v$ and at an angle $\theta$ to the horizontal is given by:</p> <p>$R = \frac{v^2}{g} \sin(2\theta)$</p> <p>where $g$ is the acceleration due to gravity.</p> <p>From this equation, we can see that the range is dependent on the sine of twice the launch angle.</p> <p>Given that the range at $15^\circ$ is $50$ m, if we launch the projectile at $45^\circ$ with the same velocity, we can compare the ranges by comparing $\sin(2 \times 15^\circ)$ and $\sin(2 \times 45^\circ)$:</p> <p>$\sin(30^\circ) = \frac{1}{2}$</p> <p>$\sin(90^\circ) = 1$</p> <p>Therefore, the range at $45^\circ$ will be twice the range at $15^\circ$, because $\sin(90^\circ)$ is twice as large as $\sin(30^\circ)$.</p> <p>So, the range when the projectile is launched at $45^\circ$ will be $2 \times 50$ m = $100$ m.</p> <p>So, $100$ m is the correct answer.</p>