suppose a soccer player boots the ball from a distance 35 m into the goal. find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be 40° above the horizontal
To find the initial speed of the ball, we need to use the kinematic equation for projectile motion. The equation is:
y = yi + viy * t - (1/2) * g * t^2
where:
y is the vertical displacement (2.4 m),
yi is the initial vertical position (0 m),
viy is the initial vertical velocity,
t is the time of flight,
g is the acceleration due to gravity (-9.8 m/s^2).
We can solve this equation to find the time of flight (t) first. Rearranging the equation, we get:
t = (viy ± sqrt(viy^2 - 4*(1/2)*g*y)) / (2*(1/2)*g)
Since the ball passes over the goal, we consider the positive value of t.
To find the initial speed (v) of the ball, we can use the horizontal range formula:
Range (R) = v * cosθ * t
where θ is the angle of projection (40°).
For the ball to just pass over the goal, the horizontal displacement (Range) should be equal to the distance to the goal (35 m). So, we have:
35 m = v * cos40° * t
Now, we can substitute the value of t from the first equation into the second equation to find v:
35 m = v * cos40° * [(viy + sqrt(viy^2 - 4*(1/2)*g*y)) / (2*(1/2)*g)]
Simplifying this equation will give us the value of v, the initial speed of the ball.
Note: The value of viy can be obtained by using the equation for vertical velocity:
viy = vi * sinθ
where vi is the initial speed.