To determine the initial velocity of the soccer ball, we can analyze its horizontal and vertical motion separately.
Given: Launch angle (θ) = 60° Horizontal distance (range) (R) = 30 m Vertical displacement (Δy) = 0 (since the ball strikes the ground)
We can use the following equations of motion:
Horizontal motion: R = Vx * t
Vertical motion: Δy = Vy * t - (1/2) * G * t²
Since the ball is launched at an angle of 60°, the initial velocity (V) can be resolved into its horizontal (Vx) and vertical (Vy) components:
Vx = V * cos(θ) Vy = V * sin(θ)
In the vertical motion equation, we can substitute the values:
0 = Vy * t - (1/2) * G * t²
We can solve this equation for t:
(1/2) * G * t² = Vy * t (1/2) * 10 m/s² * t² = V * sin(θ) * t 5t² = V * sin(θ) * t 5t = V * sin(θ)
Now, let's look at the horizontal motion equation:
R = Vx * t 30 m = V * cos(θ) * t
Since we have the value of t in terms of V and sin(θ), we can substitute that into the equation:
30 m = V * cos(θ) * (5t / V * sin(θ)) 30 m = 5t * cos(θ) / sin(θ) 30 m = 5t * (1 / tan(θ)) 30 m = 5t / √3
Now we can substitute the value of t from the previous equation:
30 m = 5 * (V * sin(θ)) / √3 30 m = (5 * V * √3) / √3 30 m = 5V
Finally, we can solve for V:
V = 30 m / 5 V = 6 m/s
Therefore, the initial velocity of the soccer ball was 6 m/s.