A model rocket is fired vertically from the ground with a constant acceleration of 44.3 m/s2 for 1.95 s at which time its fuel is exhausted. What is the maximum height (in m) reached by the rocket?

Question

Shouldn't this just be d=rt?
44.3(1.95), but I'm not getting the right distance, am I supposed to account for gravity?

Answer 1

let's assume that for the first 1.95 seconds, the given acceleration of 44.3 m/s^2 includes the effect of gravity, and for any time thereafter the effects of gravity of -4.9 m/s^2 takes over

for the first part:
a = 44.3
v = 44.3t + c
but when t = 0 (at the start)
v = 0 , ----> c = 0

v = 44.3t
s = 22.15t^2 + k
again, when t = 0 , s - 0 , so k = 0

distance covered = s = 22.15t^2 , where 0≤t≤1.95
so after 1.95 s
the height is 22.15(1.95)^2 = 84.225
and v = 44.3(1.95) = 86.385 m/s

second part:

so the effective height equation after 1.95 s:
height = -4.9t^2 + 86.385t + 84.228 , where t is the time AFTER 1.95 seconds
this is a downwards opening parabola, all we have to do is find its vertex
the t of the vertex is -b/(2a) = -86.385/-9.8 = 8.815 seconds
and the height = -4.9(8.815)^2 + 86.385(8.815) + 84.228
= 464.96 or appr 495 m

remember that the time of 8.815 is the time after the initial 1.95 seconds
so the rocket reaches the maximum of 495 m in 10.765 seconds after take-off.

better check my arithmetic.