Question

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 98 m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 31.0 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.

EXPLAIN ANSWERS PLEASE THE FIRST PERSON I ASKED BEFORE GOT IT WRONG

Answer 1

here,

writing velocity in terms of horizontal and vertical components,
Vxo = VCos53 = 98*0.60 = 58.8 m/s
Vyo = VSin53 = 98*0.79 = 77.42 m/s

the velocity at the end of 3 s = Vo + at
Vf= 98 + 31 * 3 = 191 m/s

CASE A : MAximum Altitude

lets call the maximum altitude of the rocket H and say that the height reached while the engines are running is h. Also we'll call the angle of the launch a.

we can get an expression for h as follows.

h = vsin(a)t + (1/2)at^2
h = 98 * sin(53) * 3 + 0.5*31*9
h = 232.26 + 139.5
h = 371.76 m

and we can write an expression for the time it takes the rocket to reach maximum height after the engines have failed.

t = [0 - vsin(a)]/-g
t= vsin(a)/g

then

H-h = vsin(a)t - 1/2*gt^2

if we sub in the expression for t and rearrange we get

H-h = [(vsin(a)^2]/2g

now usingvalue of h and solving H we get ,

H - 371.76 = ((191*sin53)^2 )/ (2 * 9.8 ) = 1558.92m

maximum altitude reached by the rocket is 1558.92m

Case B : Total time of flight

to obtain the total time of flight, first work out the time it takes for the rocket to fall from the top of its trajectory.

t = (v-u)/g

where v is now the final vertical velocity and u is the initial vertical velocity.

to get v we have , v^2 = u^2 + 2gH

so t = [Sqrt(2gH)]/g

now the total time is the falling time plus the time the engines are running, plus the time we calculated previously.

t = [Sqrt(2gH)]/g + 3 + vsin(a)/g

t = (Sqrt(2 * 9.8 * 1558.92))/9.8 + 3 + 191 / 9.8 = 40.32

Total time of flight is 40.32 s

Case 3: horizontal range

calculating the range is simple since the horizontal velocity stays the same for most of the flight.
We just need to work out the bit of distance traveled while the rocket accelerates for the first 3 seconds of flight.

r = ut + 1/2at^2
r = 98 * 3 + 0.5 * 31 * 9
r = 433.5 m

now we take the remaining flight time and put it in the following formula to work out the rest of the distance.

R-r = vcos(a)t

= 191 * cos(53) * 40.32 + 433.5

the horizontal range rocket projectile is 5068.15 m