A bullet is fired from a canon with velocity 500 m/s. If the angle of projection is 15° and g = 10 m/s⁻². Then the range is

The correct option is 12.5 × 10³ m.

Step-by-step derivation:

We are given the following values for a projectile motion:
Initial velocity ($u$) = 500 m/s
Angle of projection ($\theta$) = 15°
Acceleration due to gravity ($g$) = 10 m/s²

The formula for the horizontal range ($R$) of a projectile is given by:
$R=\frac{u^{2}sin\left(2\theta \right)}{g}$

Substitute the given values into the formula:
$R=\frac{{\left(500\right)}^2\times sin(2\times 15°)}{10}$

Simplify the angle and compute the square of the velocity:
$2\times 15°=30°$
${\left(500\right)}^2=\mathrm{250,000}$

Since $sin(30°)=0.5$, we can substitute this value back into the range equation:
$R=\frac{\mathrm{250,000}\times 0.5}{10}$

Calculate the final value:
$R=\frac{\mathrm{125,000}}{10}=\mathrm{12,500}\text{m}$

Expressing the result in scientific notation as presented in the options:
$R=12.5\times {10}^3\text{m}$