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=
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-> 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=
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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
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.)
or [This is the quadratic equation; , , ]
or
or
or
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.
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