A football is kicked at ground level with a speed of 18.0 m/s at an angle of 38.0 degrees to the horizontal. How much does it hit the ground?
To find out how much the football hits the ground, we need to determine the horizontal distance it travels.
We can break down the initial velocity of the football into its horizontal and vertical components. The horizontal component is given by Vx = V * cos(theta), where V is the initial speed (18.0 m/s) and theta is the angle of projection (38.0 degrees).
Vx = 18.0 m/s * cos(38.0 degrees) = 14.4 m/s
Next, we can use the equation of motion in the vertical direction to find the time it takes for the football to hit the ground. The equation is given by:
y = V * t * sin(theta) - (1/2) * g * t^2
where y is the vertical displacement (we want to find this when the ball hits the ground), V is the initial speed (18.0 m/s), t is the time, theta is the angle of projection (38.0 degrees), and g is the acceleration due to gravity (9.8 m/s^2).
Since the ball is initially at ground level, y = 0. Therefore, the equation simplifies to:
0 = V * t * sin(theta) - (1/2) * g * t^2
Rearranging the equation, we get:
(1/2) * g * t^2 = V * t * sin(theta)
Simplifying further, we have:
(1/2) * g * t = V * sin(theta)
Now, we can solve for t by dividing both sides of the equation by (1/2) * g:
t = (2 * V * sin(theta)) / g
Plugging in the values, we get:
t = (2 * 18.0 m/s * sin(38.0 degrees)) / 9.8 m/s^2
t ≈ 3.036 seconds
Finally, to find the horizontal distance (dx) the ball travels, we can multiply the horizontal component of velocity (Vx) by the time (t):
dx = Vx * t
dx = 14.4 m/s * 3.036 seconds
dx ≈ 43.8 meters
Therefore, the football hits the ground approximately 43.8 meters away from the point it was kicked.