anonymous
  • anonymous
an arrow is shot vertically upward from a platform of 25 ft high at a rate of 153 ft per second. When will the arrow hit the ground?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
So we are using gravity as the force right?
anonymous
  • anonymous
Integrate velocity to get the speed, out in the relation for the force of gravity, c is the original height, and solve
anonymous
  • anonymous
help me

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anonymous
  • anonymous
velocity is speed but with direction....
anonymous
  • anonymous
\[h=-16t ^{2}+v _{0}t+h _{0}\] this is the formula
anonymous
  • anonymous
I'm just have trouble working it out. so I wanted to see if anyone else could get an answer
anonymous
  • anonymous
Use the kinematics formulas in order to solve this with a= 9.8 m/ s^2
anonymous
  • anonymous
Yes, set that equal to zero to find the maximum height and the time qt which it occurs. Then you can find out how long it takes for the arrow to fall.
anonymous
  • anonymous
this is not that complicated. to solve this equation you will write: h=-16t^2+153t+25
anonymous
  • anonymous
You have all the variables given, except for t, which is what you're looking for.
anonymous
  • anonymous
Please post all the information in the question. Disregard my previous statements.
anonymous
  • anonymous
v_0=153ft/sec, h=0, h_0=25ft.
anonymous
  • anonymous
Just substitute and solve for t.
anonymous
  • anonymous
you guys are making this seem way more complicated. h=-16t^2+153t+25. set h to zero. you get 0=-16t^2+153t+25, then solve for t.
anonymous
  • anonymous
Yeah, maph... I said that a while ago lol.
anonymous
  • anonymous
\[0=-16t^2+153t+25 \implies t=9.7\]
anonymous
  • anonymous
I only took the positive value for t.
anonymous
  • anonymous
ok I was doing it right. Just wanted to double check. thanks for the help everyone
anonymous
  • anonymous
9.7 is rounded to the nearest 10th right?
anonymous
  • anonymous
Yes.

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