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

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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) About Me  (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) About Me  (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.
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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)
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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) About Me  (Show Source):
You can put this solution on YOUR website!
h=vt-16t%5E2 Start with the given formula


h=48t-16t%5E2 Plug in the given velocity of v=48


0=48t-16t%5E2 Plug in the height of 0 h=0 (since the height is zero at the ground)


0=-16t%5E2%2B48t Rearrange the terms.


Notice that the quadratic -16t%5E2%2B48t is in the form of At%5E2%2BBt%2BC where A=-16, B=48, and C=0


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


t+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


t+=+%28-%2848%29+%2B-+sqrt%28+%2848%29%5E2-4%28-16%29%280%29+%29%29%2F%282%28-16%29%29 Plug in A=-16, B=48, and C=0


t+=+%28-48+%2B-+sqrt%28+2304-4%28-16%29%280%29+%29%29%2F%282%28-16%29%29 Square 48 to get 2304.


t+=+%28-48+%2B-+sqrt%28+2304-0+%29%29%2F%282%28-16%29%29 Multiply 4%28-16%29%280%29 to get 0


t+=+%28-48+%2B-+sqrt%28+2304+%29%29%2F%282%28-16%29%29 Subtract 0 from 2304 to get 2304


t+=+%28-48+%2B-+sqrt%28+2304+%29%29%2F%28-32%29 Multiply 2 and -16 to get -32.


t+=+%28-48+%2B-+48%29%2F%28-32%29 Take the square root of 2304 to get 48.


t+=+%28-48+%2B+48%29%2F%28-32%29 or t+=+%28-48+-+48%29%2F%28-32%29 Break up the expression.


t+=+%280%29%2F%28-32%29 or t+=++%28-96%29%2F%28-32%29 Combine like terms.


t+=+0 or t+=+3 Simplify.


So the solutions are t+=+0 or t+=+3

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 t=3 is more interesting. So we're going to ignore t=0


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Answer:


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