A rocket is laucnhed (physics, PLEASE HELP!!!!)?

Question

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 99 m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 32.0 m/s2. At this time, its engines fail and the rocket proceeds to move as a projectile.
(a) Find the maximum altitude reached by the rocket.
(b) Find its total time of flight
(c) Find its horizontal range

I tried plugging in a few equations but i cant even seem to find A.

Answer 1

This is piecewise linear
I made an assumption that 32 m/s^2 is the net acceleration along a straight 53 degree flight path, which assumes it is net of gravity along that path.

0 For the first part of the motion it has vertical velocity as

vy(t)=sin(53)*(99+32*t)
and
y(t)=sin(53)*(99*t+16*t^2)

for t>3 until y(t)=0
we must calculate the speed at the end of the first 3 s
155.7 m/s
he altitude at this moment is
352.2 meters
so the equations for t>3 are
vy(t)=155.7-g*(t-3)
y(t)=352.2+155.7*(t-3)
-.5*g*(t-3)^2
using g=9.81
at apogee, vy(t)=0
0=155.7-g*(t-3)
t=155.7/g+3
t=18.87seconds

so the altitude is
1587.8 meters

the total flight time is found by setting y(t)=0 and solving the quadratic for the larger root

t=36.86 seconds this is the total flight time.

It is always a good idea to reconcile the existence of a second root. The second root is 0.88 seconds. This value is the equivalent time that if the rocket were given a vy0 from the ground that would bring it to apogee at the time and altitude that the rocket motor did. The root is positive since the acceleration of the rocket motor was more than 3 times g, and the angle was great enough at 53 degrees.

For range
0 vx(t)=cos(53)*(99+32*t)
x(t)=cos(53)*(99*t+16*t^2)
the horizontal speed at t=3 is

117.35 m/s
and the distance down range is
265.4 meters

for 3 x(t)=265.4+117.35*(t-3)
from above we know that impact occurs at 36.86 seconds
x(36.86)=4239 meters

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