SOLUTION: Yet another, I can't seem to make "work". If a ball is thrown straight up with an initial velocity of 48 feet per second, it's height after t seconds, is given by the formula h=

Algebra ->  Rate-of-work-word-problems -> SOLUTION: Yet another, I can't seem to make "work". If a ball is thrown straight up with an initial velocity of 48 feet per second, it's height after t seconds, is given by the formula h=      Log On


   

Question 42657This question is from textbook
: Yet another, I can't seem to make "work".
If a ball is thrown straight up with an initial velocity of 48 feet per second, it's height after t seconds, is given by the formula h= -16tsquared + 48t. How long will it take the ball to return ot its point of departure?
I have h = 0.
I did the quad formula and came out with -48 +- 72/all over 32. Reduced to -24 +- 36/2, 12/2= 6 seconds. I put it into the original formula and it doesn't work. I apologize if this isn't making sense. It's been 27 years since high school and this college algebra is getting to me. I've worked out numerous problems, I'm down to this and the other I posted. I need to understand exactly what I've done wrong.
Thank you for your time and expertise... boy I appreciate it. Best Regards, Heidi
 This question is from textbook

Found 2 solutions by psbhowmick, Nate:
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
h+=+-16t%5E2+%2B+48t

When the ball will return to ground, h = 0; you are right.
Put h = 0 in the above equation (You have commited some error in this step.)

0+=+-16t%5E2+%2B+48t
or 16t%5E2+-+48t+=+0 [This is the quadratic equation; red%28a+=+16%29, red%28b+=+-48%29, red%28c+=+0%29]
or 16t%28t-3%29+=+0
or t%28t-3%29+=+0%2F16
or t%28t-3%29+=+0

As product of two quantities = 0, so at least one of them must be zero.
Hence either t = 0 or (t - 3) = 0 i.e. either t = 0 or t = 3.
Now, at t=0 the ball was thrown up so then its height was also zero.
So t = 0 cannot be the time when it comes down.
So t = 3 must be the time when it comes down.

The ball was thrown up at time t = 0, it came down at time t = 3.
So time of flight = (time when it comes down) - (time when it was thrown) = 3 - 0 = 3.

So the answer is 3 seconds.
Got it now?

Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
h+=+-16t%5E2+%2B+48t where h is height
0+=+-16t%5E2+%2B+48t where the height from the ground is zero feet
Use Quadratic Formula:
h+=+at%5E2+%2B+bt+%2B+c
a+=+-16
b+=+48
c+=+0
t+=+%28-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
t+=+%28-48+%2B-+sqrt%28+48%5E2+-+4%2A%28-16%29%2A%280%29+%29%29%2F%282%2A%28-16%29%29+
t+=+%28-48+%2B-+sqrt%28+48%5E2+%29%29%2F%28-32%29%29+
t+=+%28-48+%2B-+48%29%2F%28-32%29%29+
t+=+%28-48+%2B+48%29%2F%28-32%29 or x+=+%28-48+-+48%29%2F%28-32%29
t+=+0 or x+=+3
+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+38%2C+-16x%5E2+%2B+48x%29+