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?
Range = Vo^2*sin(2A)/g = 15 m.
Vo^2*sin(40)/9.8 = 15
Vo^2*0.0656 = 15
Vo^2 = 228.7
Vo = 15.12 m/s.
To find the initial velocity of the soccer ball, we can break the problem down into horizontal and vertical components.
First, let's consider the horizontal component. The horizontal velocity remains constant throughout the entire motion, so we don't need to worry about it for now.
Next, let's focus on the vertical component. We can use the kinematic equation for vertical motion:
h = ut + (1/2)gt^2
Where:
h = vertical displacement (2.0 m)
u = initial vertical velocity (unknown)
t = time of flight
g = acceleration due to gravity (approximately 9.8 m/s^2)
Since the soccer ball is shot at an angle of 20 degrees with the ground, we can find the time of flight by using the horizontal component of the initial velocity. The horizontal velocity (uh) is related to the initial velocity (u) and the angle (θ) by the equation:
uh = u * cos(θ)
Similarly, the vertical velocity (uv) is related to the initial velocity (u) and the angle (θ) by the equation:
uv = u * sin(θ)
We know that the vertical displacement (h) is 2.0 m, and we can assume that the time of flight (t) is the same for both the horizontal and vertical components.
Using the equation for vertical motion, we can rewrite it as:
h = uv * t - (1/2)g * t^2
Substituting the values we know:
2.0 = (u * sin(20)) * t - (1/2) * 9.8 * t^2
Now, we have two equations:
uh = u * cos(20)
2.0 = (u * sin(20)) * t - (1/2) * 9.8 * t^2
We can solve these two equations simultaneously to find the initial velocity (u) and the time of flight (t).