The maximum range of a projectile is 22 m.

When a projectile is thrown at an angle of 15 degrees with the horizontal, the range can be calculated using the equations of projectile motion. The range is the horizontal distance covered by the projectile before it hits the ground.

To calculate the range, we can use the following equation:

Range = (v^2 * sin(2θ)) / g

Where:
- Range is the maximum horizontal distance covered by the projectile.
- v is the initial velocity of the projectile.
- θ is the launch angle.
- g is the acceleration due to gravity (approximately 9.8 m/s^2).

Calculating the range:

Given that the maximum range is 22 m, we can rearrange the equation to solve for the initial velocity:

v^2 = (Range * g) / sin(2θ)

Plugging in the values:

v^2 = (22 * 9.8) / sin(2 * 15)

v^2 = 214.6 / sin(30)

v^2 = 214.6 / 0.5

v^2 = 429.2

Taking the square root of both sides:

v = √429.2

v ≈ 20.71 m/s

Calculating the range at an angle of 15 degrees:

Now that we have the initial velocity, we can calculate the range at an angle of 15 degrees. Plugging the values into the range equation:

Range = (v^2 * sin(2θ)) / g

Range = (20.71^2 * sin(2 * 15)) / 9.8

Range = (429.2 * sin(30)) / 9.8

Range = (429.2 * 0.5) / 9.8

Range ≈ 22 m

Therefore, the range of the projectile when thrown at an angle of 15 degrees with the horizontal is also approximately 22 m, which is the maximum range.

22*10=u power2 * sin(90)
u=220power one by 2
r*10
=220*sin(30)
R= 220/20
R=11m