70 ft per sec going upward 3 sec from 6 ft high should reach 216 ft, but unfortunately gravity is pulling it and at some point it starts to go backward (at 2.176 second), so to compensate the upward slowing down and distance dropped, use -32.176 times 3 square times 1/2 = -144.792 which is the natural drop distance if the football is released mid air without a initial speed after 3 seconds. 216 - 144.792 = 71.208 ft high.
Did you leave out any part of the problem?
Hi, I have an answer for you, hope is O.K.
For this case we have that by definition, the equation of the line of the slope-intersection form is given by:

Where:
m: It's the slope
b: It is the cut-off point with the y axis
We have two points through which the line passes:

We found the slope:

Substituting we have:

Thus, the equation is of the form:

We substitute one of the points and find the cut-off point:

Finally, the equation is:

ANswer:
