a bad punter on a football team kicks a football approximately straight upward with an initial velocity of 85 ft/sec.

Question

If the ball leaves his foot from a height of 3 ft, write an equation for the vertical height s (in ft) of the ball t seconds after being kicked.

 

Find the time(s) at which the ball is at a height of 104.301563 ft. Round to 1 decimal place.

Answer 1

Part 1:

First we need the basic formula for parabolic motion, or projectile motion.  This formula is:

 

s = -.5 * g * t² + v₀ * t + s₀

 

t = time (always is the independent variable)

s = height (always is the dependent variable)

 

g = gravitational constant = 32.1740 ft/sec²

v₀ = initial velocity = 85 ft/sec (given in problem)

s₀ = initial height = 3 ft (given in problem)

 

Now we know the formula and identified all of the necessary data given in the problem.  Just plug in to write the equation for this problem:

 

s = -.5 * 32.1740 * t² + 85 * t + 3

 

Now to find the time (t) at which the ball is at a height of 104.301563 ft (s).  Plug this given height (s) into the equation and solve for time (t).  When you do this, you'll get t = 1.8 sec or 3.5 sec

Answer 2

I'm assuming you are doing this without calculus and therfore have a formula, in feet, that looks like s(t) = -16t^2 + (initial velocity)x + (initial position). So your position fumctionis

s(t) = -16t^2 + 85t + 3

 

You now want to find out what time you have to plug in to result in a position of 104.301563 ft. This leads to the equation

-16t^2 + 85t + 3 = 104.301563    solving this using the quadratic formula you get

-16t^2 + 85t - 101.301563 = 0

x = (-85 + sqrt( 85^2-4(-16)(-101.301563) )/(-32) and x = (-85-sqrt(85^2-4(-16)(-101.301563) ) /(-32)

 

These come out to 1.8 and 3.5 seconds.