A rocket runs out of fuel at an altitude of 1000 ft above ground level?

Question

A rocket is launched vertically and it runs out of fuel at an altitude of 1000 ft. above ground level, when its velocity is 78.4 ft./sec. How long will it be before the rocket again is at an altitude of 1000 ft. above ground level?


Please, please, PLEASE at least give me the equation. Setting up the equation is what I have the most trouble with. :(

Answer 1

At 1000 ft, it is going upwards at 78.4 ft/s.

Every second after running out of fuel, its speed will decrease by 32.2 ft per second (Earth's gravity). You can find how many seconds before the upward speed reaches 0 (the rocket reaches its highest altitude).

By symmetry, it will take the same time to get back down to 1000 ft altitude (and it will pass that altitude with a speed of 78.4 ft/s, downwards).


speed at time t = initial speed (upwards) - (32.2 ft/s^2)*t
s(t) = s(0) - g*t

t in seconds

You are solving for t, when s(t) = 0 and s(0) = 78.4

This is the time to reach the apex. Multiply by two to find the time it takes to fall back down to 1000 feet.

Answer 2

you don't realy have enough information to accurately answer that question, specificly (the mass of the rocket) without mass you cannot calculate inertia! but, basicly 78.4 fps per second minus 32 fps per second.

Answer 3

This is what you do multiply the Ft/sec to the 1000ft and you'll find your answer!!!