Aplacekicker must kick a football from a point 36.0 m (about 40 yards) from the goal. half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. when kicked, the ball leaves the ground with a speed of 21.6 m/s at an angle of 45.0° to the horizontal. (a) by how much does the ball clear or fall short of clearing the crossbar? (enter a negative answer if it falls short.)

Since the ball was launched at 45°above the horizontal, the 'x' and 'y' components of its initial velocity are equal, and each is 

         V₀ sin(45°)  =  V₀ cos(45°)  =  (1/2) V₀ √2  =  15.273 m/s .

The ball's horizontal speed is constant, so it reaches the goal posts in

             (36 m) / (15.273 m/s)  =  2.357 seconds.

The ball's vertical height above the ground at any time is

           H  =  -16 t² + 15.273 t            meters.

After 2.357 seconds, the ball is

                   -16 (2.357)² + 15.273 (2.357)  meters off the ground

            =     -16 (5.555)  +  15.273 (2.357)

            =        -88.88      +     35.998          meters off the ground.

This is not only well below the crossbar, it's also mathematically
below ground, down where da sun don't shan.  This is algebra's
way of telling us that the ball returned to the ground before it ever
reached the goal post, and if it ever reached the goal post at all,
it was rolling in the grass when it got there.      

Wavelength divides in half and speed stays the same.

As newton's first law states a object in motion stays in motion