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

To determine the range of a projectile fired at an angle of 45 degrees, given that its range at 15 degrees is 50 meters, we can use the properties of projectile motion. The range of a projectile depends on the initial velocity, the angle of launch, and the acceleration due to gravity. Let's break this down step by step.

Understanding Projectile Motion

In projectile motion, the range \( R \) can be calculated using the formula:

R = (v^2 * sin(2θ)) / g

where:

• v is the initial velocity of the projectile.
• θ is the angle of projection.
• g is the acceleration due to gravity (approximately 9.81 m/s²).

Calculating the Initial Velocity

First, we need to find the initial velocity \( v \) when the projectile is launched at 15 degrees. We know that the range at this angle is 50 m. Plugging the values into the range formula:

50 = (v^2 * sin(30°)) / g

Since \( sin(30°) = 0.5 \), we can rewrite the equation:

50 = (v^2 * 0.5) / g

Rearranging gives us:

v^2 = 100g

Substituting \( g \) with 9.81 m/s²:

v^2 = 100 * 9.81 = 981

Thus, the initial velocity \( v \) is:

v = √981 ≈ 31.34 m/s

Finding the Range at 45 Degrees

Now that we have the initial velocity, we can calculate the range when the projectile is fired at 45 degrees:

R = (v^2 * sin(90°)) / g

Since \( sin(90°) = 1 \), the formula simplifies to:

R = (v^2) / g

Substituting \( v^2 \) and \( g \):

R = 981 / 9.81 ≈ 100 m

Final Result

Therefore, if the projectile is fired with the same speed at an angle of 45 degrees, its range will be approximately 100 meters. This demonstrates how the angle of projection significantly influences the range of a projectile, with 45 degrees yielding the maximum range for a given initial velocity in ideal conditions.