(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.20 m/s^2. When it is 260 m above the launch pad, it discards a used fuel canister by simply disconnecting it. Once it is disconnected, the only force acting on the canister is gravity (air resistance can be ignored).

How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration?

2. Relevant equations

1. v^2 = v(0)^2 - 2g(y - y(0))

2. y = y(0) + v(0)t - (1/2)at^2

3. v = v(0) - gt

3. The attempt at a solution

first i wanted to find the speed of the rocket at 260m by using equation 1. and solving for v. I attained 40.8m/s.

then I used this 40.8m/s as the initial velocity of the canister going up and v = 0 (speed of the canister at its maximum height) as the final velocity to find the initial height of the canister before it begins to drop. I attained 345m.

once i found this i wanted to find the final velocity of the canister before it hit the pad and used equation 1 again. I attained v = 82.2m/s.

after, i found the time it took the canister to hit the ground by using equation 3. I attained 8.39s.

finally, i used this time interval along with v(0) = 40.8m/s, y(0) = 260m, a = 3.2m/s^2 in equation 2. I attained 715m which is wrong according to the course website.

can anybody help me?

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# Homework Help: Height of a rocket at a certain time

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