SOLUTION: Ths is a summer math problem I was assigned to complete. A punter can kick a football with an initial velocity of 48 feet per second. How many seconds will it take for the ball to

Question 211170: Ths is a summer math problem I was assigned to complete. A punter can kick a football with an initial velocity of 48 feet per second. How many seconds will it take for the ball to return to the ground?(Hint: Use the formula h=vt-16t2*)I was unsure whether or not this formula was correct for some reason I thought the formula was h=v-16t2*. I still need help wth completing the problem though. Thank you in adance the sooner this is answered the better.
*The 2 in the problems above is a square.
Found 3 solutions by nerdybill, Alan3354, jim_thompson5910:
Answer by nerdybill(7384)   (Show Source): You can put this solution on YOUR website!
A punter can kick a football with an initial velocity of 48 feet per second. How many seconds will it take for the ball to return to the ground?(Hint: Use the formula h=vt-16t2*)
.
No, h=vt-16t^2
is correct
since v (velocity) is feet/sec and if you didn't multiply with time your units would not work.
.
Since h = height (in feet)
set to zero to find out when it hits the ground
.
0 = vt-16t^2
plugging in the given initial velocity:
0 = 48t-16t^2
0 = 3t-t^2
0 = t(3-t)
t = {0, 3}
.
Well, 0 secs is "before" the punter kicked the ball so we're left with:
t = 3 seconds

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
A punter can kick a football with an initial velocity of 48 feet per second. How many seconds will it take for the ball to return to the ground?(Hint: Use the formula h=vt-16t2*)I was unsure whether or not this formula was correct for some reason I thought the formula was h=v-16t2*.
-----------
It can't be h=v-16t^2, because the units don't match.
h = distance and v is speed. You can't add and subtract feet and feet/sec.
Multiplying v*t gives feet/sec x 1/sec = feet.
----------------
h=48t-16t^2
The ground is h = 0
48t - 16t^2 = 0
t*(48 - 16t) = 0
t = 0 (at the start)
t = 3 (return to the ground)
--------------
BTW, the -16t^2 term is acceleration due to gravity. It's (gt^2)/2, where g is -32 ft/sec/sec.
Muliplying ft/sec/sec by sec^2 gives feet, so the units are uniform.
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Start with the given formula

Plug in the given velocity of

Plug in the height of 0 (since the height is zero at the ground)

Rearrange the terms.

Notice that the quadratic is in the form of where , , and

Let's use the quadratic formula to solve for "t":

Start with the quadratic formula

Plug in , , and

Square to get .

Multiply to get

Subtract from to get

Multiply and to get .

Take the square root of to get .

or Break up the expression.

or Combine like terms.

or Simplify.

So the solutions are or

This means that at times of 0 and 3 seconds, the ball will be at the ground.

Since we already know that the ball is at the ground at 0 seconds, the other solution is more interesting. So we're going to ignore

====================================================

Answer:

So it will take 3 seconds for the ball to return to the ground.
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