how long does it take to rise to 250m?
s = 1/2 at^2, so
3.45/2 t^2 = 250
t = 12 seconds
At that point,
v = at = 3.45*12 = 41.4 m/s
How long does it take for the canister to fall from that height, given its upward speed?
h(t) = 250 + 41.4t - 4.9t^2
h=0 after another 12.52 seconds
Now the rocket's height at that time is
h = 250 + 41.4t + 1.725 t^2
where t=12.52
h = 1038.72
To find the canister's maximum height, just find the vertex of
250 + 41.4t - 4.9t^2
That is 337.45m
so, it traveled 337.45*2 - 250 = 424.90 m after release.
As always, check my math.
During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.45 m/s2 . When it is 250 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).
1) How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration? Answer in meters (m).
2) What total distance did the canister travel between its release and its crash onto the launch pad? Answer in meters (m).
1 answer