Quadratic Equation

Question

"Ms. Powers can kick a soccer ball up into the air with an initial velocity of 44 feet per second. Using the formula h=vt-16t^2, where h=height above the ground, v=initial velocity, and t=time in seconds, find how many seconds it will take for the ball to return to the ground."

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Answer 1

You can use the given equation for the upward motion of the ball to its pinnacle height, and you can use the same formula for the travel back to the ground.

 

Upward: h = vt - 16t^2

 

where v = 44 feet/s, h is the distance traveled from the ground to the maximum height, and t is the time of upward travel.

 

Downward: h = 0*t + 16t^2

 

For downward travel, the initial velocity is 0, since the ball starts its downward motion from rest, and the sign of the quadratic term is changed from a minus to a plus. In going up, the ball decelerates, so the quadratic term is negative. But, in going back down, the ball accelerates.

 

Using symmetry, we can be assured that the time going up equals the time going down. So, we have two equations and two unknowns, h and t:

 

h = vt - 16t^2

h = 16t^2

 

Let's subtract the second equation from the first. Then, we get

 

0 = vt - 16t^2 - 16t^2

 

We can then write

 

32t^2 = vt

 

We can cancel a t on both sides of the last equation.

 

32t = v

 

or

 

t = v/32 = 44/32 = 1.375 sec

 

But, remember, t represents the time going up or the time going down. The total time of flight is time up plus time down, or t + t = 2t = 2.75 sec