A soccer ball was kicked off a cliff that was h = 8.4 m above the ground. The ball was kicked at a speed of 25 m s-1 and an angle of 20 to the horizon.
What is magnitude of the horizontal displacement of the ball? Use g = 9.8 m s-2.
proper equation and help plz
See previous post: Thu, 12-15-16, 9:33 PM.
To find the magnitude of the horizontal displacement of the ball, we can use the equations of projectile motion. The horizontal displacement is the distance the ball travels horizontally from its initial position to its final position.
The horizontal component of the initial velocity can be calculated using the angle and initial speed. The formula to find the horizontal component is:
Vx = V * cos(theta)
where Vx is the horizontal component of the velocity, V is the initial speed of the ball, and theta is the angle to the horizon.
Substituting in the given values:
Vx = 25 m/s * cos(20)
Next, we need to find the time taken by the ball to reach the ground. We can use the equation:
h = (1/2) * g * t^2
where h is the vertical distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken.
Rearranging the equation to solve for t:
t = sqrt(2h/g)
Substituting the given value of h = 8.4 m:
t = sqrt(2 * 8.4 / 9.8)
Now that we have the time taken, we can find the horizontal displacement using the formula:
Horizontal displacement = Vx * t
Substituting the previously calculated values:
Horizontal displacement = (25 m/s * cos(20)) * sqrt(2 * 8.4 / 9.8)
Calculating this expression will give you the magnitude of the horizontal displacement of the ball.