I know this has been posted many times but I keep trying and cannot get the right answer, A place kicker must kick a football from a point 39.9 m from a goal. As a result of the kick, the ball must clear the crossbar, which is 3.05 m high. When kicked the ball leaves the ground with a speed of 20.6 m/s at an angle of 53° to the horizontal. By how much does the ball clear or fall short of clearing the crossbar?

I got that the time was 3.35 and because v=d/t I got 11.91 and then made a triangle of 20.6 aas the hyp and 11.91 at the bottom so then i got 16.81 which crosses the bar by 13.75 but that's not the right answer, PLEASE HELP

To solve this problem, we can break it down into two components: the vertical component and the horizontal component of the kick.

First, let's find the time it takes for the ball to reach its maximum height. We can use the vertical component of the velocity and the acceleration due to gravity.

Using the equation v = u + at, where v is the final vertical velocity, u is the initial vertical velocity, a is the acceleration, and t is the time, we can solve for t.

Given:
u = 20.6 m/s (initial vertical velocity)
a = -9.8 m/s^2 (acceleration due to gravity)
v = 0 m/s (final vertical velocity at maximum height)

Rearranging the equation, we get:
t = (v - u) / a

Substituting the values, we have:
t = (0 - 20.6) / -9.8
t ≈ 2.1 seconds

Next, let's find the maximum height reached by the ball. We can use the formula:
s = ut + (1/2)at^2

Given:
u = 20.6 m/s (initial vertical velocity)
a = -9.8 m/s^2 (acceleration due to gravity)
t = 2.1 seconds (time taken to reach maximum height)

Substituting the values, we have:
s = 20.6 * 2.1 + (1/2) * (-9.8) * (2.1)^2
s ≈ 21.7 meters

Now let's find the horizontal distance covered by the ball. We can use the horizontal component of the velocity and the time of flight.

Using the equation d = vt, where d is the horizontal distance covered, v is the horizontal velocity, and t is the time of flight, we can solve for d.

Given:
v = 20.6 m/s * cos(53°) (horizontal component of the velocity)
t = 2 * 2.1 (assuming symmetry in time)

Substituting the values, we have:
d = 20.6 * cos(53°) * 2 * 2.1
d ≈ 42.7 meters

Finally, let's determine how much the ball clears or falls short of clearing the crossbar. We can subtract the crossbar height from the maximum height.

Given:
Crossbar height = 3.05 meters
Maximum height = 21.7 meters

Subtracting the crossbar height from the maximum height, we have:
Height clearance = 21.7 - 3.05
Height clearance ≈ 18.65 meters

Therefore, the ball clears the crossbar by approximately 18.65 meters.