A projectile thrown with an initial speed u and angle of projection 15° to the horizontal has a range R. If the same projectile is thrown at an angle of 45° to the horizontal with speed 2u, its range will be

Correct Option: The correct answer is 8 R.

Let's derive the step-by-step solution to find the range of the projectile in the second case.

The horizontal range of a projectile projected with speed $u$ at an angle $\theta$ to the horizontal is given by the formula:
$R=\frac{u^{2}sin\left(2\theta \right)}{g}$
where $g$ is the acceleration due to gravity.

Case 1:
The projectile is thrown with initial speed $u$ at an angle ${\theta }_1=15°$, and its range is $R$.
Substituting these values into the range formula:
$R=\frac{u^{2}sin(2\times 15°)}{g}$
$R=\frac{u^{2}sin(30°)}{g}$

Since $sin(30°)=\frac{1}{2}$, we have:
$R=\frac{u^2}{2g}$
Rearranging this gives:
$\frac{u^2}{g}=2R$ --- (Equation 1)

Case 2:
The same projectile is thrown with speed $u_2=2u$ at an angle ${\theta }_2=45°$.
Let the new range be $R_2$:
$R_2=\frac{{\left(2u\right)}^{2}sin(2\times 45°)}{g}$
$R_2=\frac{4u^{2}sin(90°)}{g}$

Since $sin(90°)=1$, we have:
$R_2=\frac{4u^2}{g}$

Now, substituting the value of $\frac{u^2}{g}$ from Equation 1:
$R_2=4\times \left(2R\right)$
$R_2=8R$

Thus, the range of the projectile will be 8 R.