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?
2(3.00)(cos60)/9.80
(3.00)(cos60)(.530219635)
0.795m
The looks like a pretty short punt to me: about 2 feet. The correct formula is (horizontal velocity component)* (time of flight). There should be a sin 60 and a v^2 in the answer, besides the factors already there in your first equation.
To find the horizontal distance traveled by the football, we can use the projectile motion equations.
The horizontal and vertical components of the initial velocity can be calculated using trigonometry. With an initial velocity of 3.00 m/s and an angle of 60.0°, the horizontal component is given by Vx = V * cosθ, where V is the initial velocity and θ is the angle above the horizontal.
In this case, Vx = 3.00 * cos(60.0°) = 3.00 * 0.5 = 1.50 m/s.
To find the time of flight, we need to consider the vertical motion. The formula to find the time of flight is given by t = 2 * (V * sinθ) / g, where g is the acceleration due to gravity (9.80 m/s^2).
In this case, t = 2 * (3.00 * sin(60.0°)) / 9.80 = 2 * (3.00 * 0.866) / 9.80 = 0.346 seconds.
Now, we can find the horizontal distance traveled by multiplying the time of flight by the horizontal component of velocity.
Distance = Vx * t = 1.50 m/s * 0.346 s = 0.519 m.
Therefore, the horizontal distance traveled by the football is approximately 0.519 meters.