A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 76.8 m/s2 for 1.99 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?

Explanation:

Take note that when the fuel of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be moving up until all the forces of gravity would dominate and change the direction of the rocket. Hence, there will be a need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction.

Given:

a = 76.8 m/s^2

t = 1.99 s

Solution:

d = vi (t) + 0.5 (a) (t^2)

d = (0) (1.99) + 0.5 (76.8) (1.99)^2

d = 0+38.4×3.9601

d = 152.068m

vf = vi + at

vf = 0 + 76.8 (1.99)

vf = 152.83 m/s (velocity when the fuel is consumed completely)

Then, we calculate the time it takes until it reaches the maximum height.

vf = vi + at

0 = 152.83+(-9.8) (t)

0 = 152.83 + (-9.8) (t)

-152.83 = -9.8t

t = 152.83/9.8 s

t = 15.59s

Then, the second distance

d= vi (t) + 0.5 (a) (t^2)

d = 152.83 (15.59) + 0.5 (-9.8) (15.59^2)

d = 2382.6197- 1190.93

d = 1191.68m

Then, we determine the maximum altitude:

d1 + d2 = 152.068 m + 1191.68m = 1343.748m