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?

Question

anonymous · Answer

So we are using gravity as the force right?

anonymous · Answer

Integrate velocity to get the speed, out in the relation for the force of gravity, c is the original height, and solve

anonymous · Answer

help me

anonymous · Answer

velocity is speed but with direction....

anonymous · Answer

$$h=-16t ^{2}+v _{0}t+h _{0}$$ this is the formula

anonymous · Answer

I'm just have trouble working it out. so I wanted to see if anyone else could get an answer

anonymous · Answer

Use the kinematics formulas in order to solve this with a= 9.8 m/ s^2

anonymous · Answer

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 · Answer

this is not that complicated.  to solve this equation you will write: h=-16t^2+153t+25

anonymous · Answer

You have all the variables given, except for t, which is what you're looking for.

anonymous · Answer

Please post all the information in the question. Disregard my previous statements.

anonymous · Answer

v_0=153ft/sec, h=0, h_0=25ft.

anonymous · Answer

Just substitute and solve for t.

anonymous · Answer

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 · Answer

Yeah, maph... I said that a while ago lol.

anonymous · Answer

$$0=-16t^2+153t+25 \implies t=9.7$$

anonymous · Answer

I only took the positive value for t.

anonymous · Answer

ok I was doing it right. Just wanted to double check. thanks for the help everyone

anonymous · Answer

9.7 is rounded to the nearest 10th right?

anonymous · Answer

Yes.