A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 88.6 m/s2 for 1.57 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

Dear student,

ok 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 = 86 m/s^2
t = 1.7 seconds

Step 1: Find the distance to the point where fuel runs out
d = vi(t) + .5(a)(t^2)
d = (0)(1.7) + .5(86)(1.7)^2
d1 = 124.27m

Step 2: Find the velocity at the point where the fuel runs out
vf = vi + at
vf = 0 + 86(1.7)
vf = 146.2 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 = 146.2 + (-9.8)(t)
t = 14.92

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 = 146.2(14.92) + .5(-9.8)(14.92^2)
d2 = 1 090.53

Step 5: Add our distances to find maximum altitude:
so our total distance above the ground is d1 + d2 =
124.27m + 1090.53m = 1 214.8 m

Eevee,