The engine of a toy rocket supplies an average acceleration of 38.0 m/s/s to the rocket for an interval of 0.80 s. A.)If the toy roocket is launched vertically, how high does it rise in this interval? B.) How fast is the rocket moving at the end of 0.80 s? C.) What altitude does the rocket reach before falling back to Earth? D.) How long does it take the rocket to reach this altitude?

Answer 1

A) Solve Y = (a/2) T^2.
a is the acceleration and T is the 0.80 s rocket burn time.

B) Solve
V = a T, using the T = 0.8 s
V is the rocket velocity at 0.8 s.

C) Add Y (from part (A) ) and the additional distance that it rises after burnout, Y'.
After burnout, it continues to rise until (1/2) V^2 = g Y'.
Solve for the additiona altitude gain, Y'. The total altitude at that time is Y' + Y

D) 0.8 s PLUS the time it rises after burnout, T'.
g T' = V
T' = V/g

Total time rising = T + T'.

Answer 2

16.12