To find the maximum horizontal range covered by the projectile, we need to determine the time it takes for the projectile to reach the ground. We can break down the initial velocity into horizontal and vertical components:
Initial velocity (u) = 25 m/s Angle of projection (θ) = 60°
Horizontal component of velocity (u_x) = u * cos(θ) Vertical component of velocity (u_y) = u * sin(θ)
u_x = 25 * cos(60°) = 25 * 0.5 = 12.5 m/s u_y = 25 * sin(60°) = 25 * √3/2 ≈ 21.65 m/s
The time of flight (T) can be determined using the vertical component of motion. At the maximum height, the vertical component of velocity becomes zero.
Using the equation v = u + at, where v = 0, u = u_y, and a = -g (acceleration due to gravity):
0 = u_y - g * t
t = u_y / g = 21.65 / 10 ≈ 2.165 s
The horizontal range (R) can be calculated using the horizontal component of velocity and the time of flight:
R = u_x * T = 12.5 * 2.165 ≈ 27.0825 m
Therefore, the maximum horizontal range covered by the projectile is approximately 27.0825 meters.