A man kicks a ball at an angle of 55 degrees above the ground with an initial velocity of 25 m/s. How far and how high does the ball travel?
To determine how far and how high the ball travels, we can break down the motion into horizontal and vertical components.
First, let's analyze the vertical component. The initial velocity in the vertical direction is given by V0y = V0 * sin(θ), where V0 is the initial velocity and θ is the angle of projection.
V0y = 25 m/s * sin(55°)
V0y = 25 m/s * 0.819
V0y ≈ 20.475 m/s
The vertical motion of the ball can be described by the equation:
y = V0y * t - 0.5 * g * t^2
Since we want to find the highest point reached by the ball, we can determine the time it takes to reach its highest point using the formula:
V0y = g * t
t = V0y / g
t = 20.475 m/s / 9.8 m/s^2
t ≈ 2.089 seconds
Now, we can substitute the time into the equation for y to find the highest point:
y = (20.475 m/s) * (2.089 s) - 0.5 * (9.8 m/s^2) * (2.089 s)^2
y ≈ 21.366 meters
Therefore, the ball reaches a height of approximately 21.366 meters.
Next, let's analyze the horizontal component. The initial velocity in the horizontal direction is given by V0x = V0 * cos(θ), where V0 is the initial velocity and θ is the angle of projection.
V0x = 25 m/s * cos(55°)
V0x = 25 m/s * 0.574
V0x ≈ 14.35 m/s
Since there is no acceleration in the horizontal direction, we can use the equation for distance: d = V0x * t, where d is the horizontal distance and t is the time.
d = (14.35 m/s) * (2.089 s)
d ≈ 29.91 meters
Therefore, the ball travels a horizontal distance of approximately 29.91 meters.
In summary, the ball reaches a height of approximately 21.366 meters and travels a horizontal distance of approximately 29.91 meters.