A kicker punts a football from 3 feet above the ground with an initial velocity of 47 feet per second.

Question

1. Write an equation that gives the height (in feet) of the football as a function of the time (in secs) since it was punted.

Okay. According to my text book, h = -16t^2 + vt + s (T= secs, V= vertical velocity, S= initial height)

So, I entered the info into the equation and got
h =-16t^2 + 47t +3

That's what they're looking for, right?

2. Find the height (in feet) of the football 2 seconds after the punt.

So, I entered 2 in the equation and got
h = -16(2)^2 + 47(2) + 3
h = -32^2 + 97 = -927ft

But how can the height be negative?
I must have done something wrong.

3. Calculate how many seconds after the punt the ball would hit the ground.
Now I'm lost.

Any help you can provide would be great. Thanks!

Answer 1

Ahem.
h = -16(2)^2 + 47(2) + 3

You do not have -(16*2)^2, you just have -16 * 2^2 = -16*4

So,

h(2) = -64 + 97 = 33

Ball hits the ground when h=0, so

-16t^2 + 47t + 3 = 0
-(t-3)(16t+1) = 0
t = 3

Answer 2

Thanks.