Double check this for me:
A player kicks a football with an initial velocity of 3.00 m/s at an angle of 60.0* above the horizontal. What is the horizontal distance traveled by the football?
s = vi*t + (1/2)(a)(t2)
0 = 3sin(30)t + 0.5(-9.8)(t2)
t = 0.306 seconds
x = vt there is no horizontal acceleration
x = 3cos(30)(0.306 seconds)
x = 0.795011321 meters
x = 0.795 meters
That would not even get over the line of scrimmage for a PAT! You DID correctly solve the quadratic equation for t. You should have used sin 60 degrees in the s=0 equation and cos 60 degrees in the x equation. You got the right answer anyway, because sin (60) cos(60) appears in the answer,. and that equals sin(30) cos(30).
The horizontal distance travelled is
X = (V^2/g) sin 2A, where A = 60 degrees.
The problem with your low answer is that they (or you) probably made a typo error in the initial velocity of the kicked football. It is more likely to be 30 m/s.
To double-check the calculation, let's go through the steps again:
First, we can break down the initial velocity into its vertical and horizontal components. The vertical component can be calculated using the equation v_y = v * sin(theta), where v is the initial velocity (3.00 m/s) and theta is the angle (60.0 degrees). Thus, v_y = 3.00 m/s * sin(60.0) = 2.60 m/s.
Next, we can calculate the time it takes for the football to reach its highest point. Using the equation v_yf = v_yi + a_y * t, where v_yf is the final vertical velocity (0 m/s), v_yi is the initial vertical velocity (2.60 m/s), a_y is the vertical acceleration (-9.8 m/s^2), and t is the time. Rearranging the equation, we get t = (v_yf - v_yi) / a_y = (-2.60 m/s) / (-9.8 m/s^2) = 0.265 seconds.
Since the football reaches its maximum height halfway through its total flight time, the total time of flight can be calculated as 2 * t = 2 * 0.265 seconds = 0.530 seconds.
Finally, we can calculate the horizontal distance traveled by the football using the equation x = v_x * t, where v_x is the horizontal component of the initial velocity and t is the time of flight. The horizontal component, v_x, can be found using the equation v_x = v * cos(theta), where v is the initial velocity (3.00 m/s) and theta is the angle (60.0 degrees). Thus, v_x = 3.00 m/s * cos(60.0) = 1.50 m/s. Plugging in the values, we get x = (1.50 m/s) * (0.530 seconds) = 0.795 meters.
Therefore, the horizontal distance traveled by the football is indeed 0.795 meters.