Question

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 91.3 m/s2 for 1.79 seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?

Answer 1

when the fuel runs out, acceleration = 0.. however the rocket is still going up until the forces of gravity finally make it change direction.

so we must find 2 distances, one from launchpad till fuel runs out, then from the point where fuel runs out to our highest altitude.

a = 91.3 m/s^2
t = 1.79 seconds

Step 1: Find the distance to the point where fuel runs out
d = vi(t) + .5(a)(t^2)
d = (0)(1.79) + .5(91.3)(1.79)^2
d1 = 146.26m

Step 2: Find the velocity at the point where the fuel runs out
vf = vi + at
vf = 0 + 91.3(1.79)
vf = 163.427 m/s (this is our velocity when the fuel runs out)

Step 3: Find the time of our new distance equation (vf = 0 = maximum altitude)
(hint: at this point, gravity kicks in because the rocket stops accelerating)
vf = vi + at
0 = 163.427 + (-9.8)(t)
t = 16.676s

Step 4: Find the distance up until the point where vf = 0 or t =15 seconds (maximum altitude before the rocket switches direction)

d= vi(t) + .5(a)(t^2)
d = 163.427(16.676) + .5(-9.8)(16.676^2) =2725.345-1362.63
d2 = 1362.70m

Step 5: Add our distances to find maximum altitude:
so our total distance above the ground is d1 + d2 =
146.26m + 1362.70m = 1508.96 m