Question on Physics problem of time and distance

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Homework Help Overview

The problem involves a ball thrown from the top of a 96-foot building with an initial velocity of 80 ft/s, and it seeks to determine the time it takes for the ball to reach the ground and the top of the building using the equation d = 96 + 80t - 16t².

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to solve for the time it takes for the ball to reach the ground and the top of the building using the provided quadratic equation. Some participants suggest substituting values into the equation to find the corresponding times.

Discussion Status

Participants are engaging with the problem, with one providing a welcome and suggesting the use of the quadratic formula. The original poster reports finding specific times for the two scenarios, while another participant notes an additional solution that raises questions about the physical interpretation of negative time.

Contextual Notes

There is a mention of the quadratic formula and the implications of negative time in the context of the problem, indicating a need to consider the physical meaning of the solutions obtained.

Daniol10
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I have this problem I am doing for college algebra. Its been giving me a lot of trouble and i have know idea how to handle it. the problem is that is a ball that were thrown from the top of a 96 foot building traveling at an initial velocity of 80 ft/s would travel from there all the way to the ground how many seconds would it reach the ground(a) and the top of the building(b)? they also give an equation for that represents the distance traveled (d=96+80t-16t^2) I've been stuck on the problem for a good while and have now idea how to go by it
 
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Welcome to PF!

Hi Daniol10! Welcome to PF! :smile:

(try using the X2 button just above the Reply box :wink:)

ok, so you have an equation d = 96 + 80t - 16t2,

and you need to sove it for (a) d = 0, and (b) d = 96.

sooo … just put that value of d into the equation, and solve it as an ordinary quadratic equation in t …

what do you get? :smile:
 
Thanks! I had forgotten about the quadratic formula. I ended up with 6 sec for (a) and 5 sec for (b)
 
Fine! :smile:

btw, you'll notice that both equations have another solution …

eg for (a) it's -1 s, which is when it would have had to be thrown from the ground to follow the same trajectory! :wink:
 

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