Projectile Motion Question?

Question

A rocket is launched at an angle of 50.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 28.0 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.

Answer 1

This is piecewise linear
I made an assumption that 28 m/s^2 is the net acceleration along a straight 50 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(50)*(103+28*t)
and
y(t)=sin(50)*(103*t+14*t^2)

for t>3 until y(t)=0
we must calculate the speed at the end of the first 3 s
143.25 m/s
he altitude at this moment is
333.23 meters
so the equations for t>3 are
vy(t)=143.25-g*(t-3)
y(t)=333.23+143.25*(t-3)
-.5*g*(t-3)^2
using g=9.81
at apogee, vy(t)=0
0=143.25-g*(t-3)
t=143.25/g+3
t=17.6 seconds

so the altitude is
1373 meters

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

t=34.37 seconds this is the total flight time.


For range
0 vx(t)=cos(50)*(103+28*t)
x(t)=cos(50)*(103*t+14*t^2)
the horizontal speed at t=3 is

120.2 m/s
and the distance down range is
279.6 meters

for 3 x(t)=279.6+120.2*(t-3)
from above we know that impact occurs at 34.37seconds
x(34.37)=4050 meters

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