a soccer player shoots a soccer ball at the opposing goal from 15 m away. If the ball leaves his foot at an angle of 20(degree) with the ground, what must its initial velocity be if it enter the goal 2.0 m above the ground?
To find the initial velocity of the soccer ball, we can use the principles of projectile motion. First, let's break down the given information:
- Distance from the soccer player to the opposing goal (horizontal distance) = 15 m
- Angle of the ball with the ground = 20 degrees
- Height of the goal above the ground = 2.0 m
Now, let's determine the components of the initial velocity of the soccer ball. Since there is no acceleration in the horizontal direction, the horizontal velocity remains constant. Therefore, we can use the horizontal distance to find the time of flight.
1. Horizontal component of initial velocity (Vx):
The horizontal component of the initial velocity can be found using the formula:
Vx = V * cos(theta)
where V is the initial velocity of the ball and theta is the angle with the ground.
2. Vertical component of initial velocity (Vy):
The vertical component of the initial velocity can be found using the formula:
Vy = V * sin(theta)
3. Time of flight (t):
The time of flight can be found using the formula:
t = 2 * (Vy / g)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
4. Vertical displacement (d):
The vertical displacement can be found using the formula:
d = (Vy * t) + (0.5 * g * t^2)
Since the ball enters the goal 2.0 m above the ground, the vertical displacement (d) should be equal to 2.0 m.
Using these equations, we can solve for the initial velocity (V). Let's follow the steps:
Step 1: Find Vx:
Vx = V * cos(theta)
Vx = V * cos(20)
Step 2: Find Vy:
Vy = V * sin(theta)
Vy = V * sin(20)
Step 3: Find t:
t = 2 * (Vy / g)
Step 4: Find d:
d = (Vy * t) + (0.5 * g * t^2)
Now, substitute the values into the equation and solve for V.