The horizontal range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be

To find the horizontal range of the projectile when fired at an angle of 45°, we can use the formula for the horizontal range of a projectile.

The horizontal range is given by the formula:
$R=\frac{u^{2}sin\left(2\theta \right)}{g}$
where:
• u is the initial speed of projection,
• θ is the angle of projection with the horizontal,
• g is the acceleration due to gravity.

Step 1: Analyze the first case
Given:
• Initial angle of projection, ${\theta }_1=15°$
• Horizontal range, $R_1=50\text{ m}$

Substitute these values into the range formula:
$50=\frac{u^{2}sin(2\times 15°)}{g}$
$50=\frac{u^{2}sin(30°)}{g}$

Since we know that $sin(30°)=\frac{1}{2}$, we can write:
$50=\frac{u^2}{g}\times \frac{1}{2}$
$\frac{u^2}{g}=100$

Step 2: Analyze the second case
Given:
• Same initial speed u,
• New angle of projection, ${\theta }_2=45°$

Let the new horizontal range be $R_2$. Using the formula:
$R_2=\frac{u^{2}sin(2\times 45°)}{g}$
$R_2=\frac{u^{2}sin(90°)}{g}$

Since $sin(90°)=1$, the equation simplifies to:
$R_2=\frac{u^2}{g}$

Substituting the value of $\frac{u^2}{g}$ from Step 1:
$R_2=100\text{ m}$

Therefore, the horizontal range when fired at an angle of 45° is 100 m.