Good problem that will leave many confused (i mean it )?

Question

A rocket is launched at an angle of 42 degrees above
the horizontal with an initial speed of 77 m/s,
as shown. It moves for 28 s along its initial

line of motion with an overall acceleration of
25 m/s2. At this time its engines fail and therocket proceeds to move as a free body.
The acceleration of gravity is 9.81 m/s2 :

a) What is the rocket's maximum altitude?
Answer in units of m.

b)What is the rocket's total time of flight?
Answer in units of s.

c) What is the rocket's horizontal range? An-
swer in units of m.

Answer 1

Oh come on!! This is simple, 1'st year, 2-D, Newtonian dynamics and kinematics.

First, find out how far the rocket travels under power.
s = V0*t + at²/2 = 77*28 + 25*28²/2 = 11956 m (along its original flight path) At this point, it is
11956*sin(42) = 8000 meters up and
11956*cos(42) = 8885 meters downrange. It is traveling at a velocity of v = V0 + at = 77 + 25*28 = 777 m/s along it's original trajectory so its vertical component is
Vv = 777*sin(42) = 519.9 m/s and it's horizontal component is
Vh = 777*cos*42) = 577.42 m/s.
The rocket reaches its maximum height when its upward velocity is 0 and
V² = V0² - 2gh => h = V0²/2g = 519.9²/2*9.8 = 13791 m
BUT!! that's 13791 m -above- the point it went ballistic so the total altitude is 13791 + 8000 = 21791 m

How long is it in flight? Well, we know for sure about the first 28 seconds ☺ How long did it take to go that last 13791 m? 13791 = gt²/2 so √(2*13791/9.8) = t = 53.05 s.
Then how long did it take to fall the -total- 21791 m from apogee? Again, 21791 = gt²/2 => t = 66.69 s. So the total time of flight is 28+53.05+66.69 = 147.74 s.

How far downrange did it land? Well...... It started 8885 m downrange with a horizontal component of 577.42 m/s
and it traveled for 53.05+66.69 = 119.74 s before it impacted, so it went 577.42*119.74 + 8885 = 78025 m.

See how easy that was? And aren't you just *so* glad that they didn't throw in 'real world' stuff such as aerodynamic effects, acceleration variance (due to the mass of the rocket changing as its fuel burns), Coriolis force, the fact that the Earth really *isn't* flat, gravitational variations due to geographic location *and* altitude, etc. etc. ? ☺


Doug
(Who has had some slight experience with both rocketry and ballistics.)

Answer 2

Hi,
You have to split this problem into two parts:
1) The Rocket's motion under the influence of the 25 m/s^2
accelaration.
AND
2) The rest of its flightpath.

So,
1) Initial velocity of rocket u = 77 m/s
Total acc. a = 25 m/s^2
Time Taken t = 28 s
and,
Final velocity:
v = u + at
= 77 + 25*28
= 777 m/s
In this time t,
Horizontal distance covered:
d1 = (ut + 1/2at^2) * cos42
[considering horizontal components of
acc. and velocity]

=> d1 = 31556 * cos42 m
= 23450.68 m [assuming cos42 = 0.7431]

And,
Vertical Distance covered:
s1 = (ut + 1/2at^2) * sin42
[similar,but I am neglecting 'g' as you have
mentioned 'overall' accelaration=25 m/s^2]

=>s1 = 31556 * sin42 m
= 21115.09 m [assuming sin42 = 0.6691]

2) Now the rocket is at a horiz. displacement of d1 and
a vertical displacement s1 from launchpoint travelling
at a velocity v = 777m/s , angle of 42 degrees,and
no force acting on it except gravity 'g'.

So,
The crux now is to find the time T taken by the rocket
to make its way to the ground.

Accelaration on the Rocket:
g = - 9.8 m/s^2 [ -ive because 'g' is downwards]

Initial velocity of the Rocket:
u = + vsin42 [+ive bcoz velocity is upwards]
=>u = 777 * sin42
= 519.91 m/s [vertical component of vel.]

And Distance to be covered(vertical):
s1 = - 21115.09 [-ive bcoz the rocket will go
downwards]

Using the equation,
s1 = ut + 1/2gt^2,
=> -21115.09 = 519.91*t - 1/2*9.81*t^2

If you solve this quadratic,you get
t = 137.2 s
That is the time the rocket takes to
reach the ground.

In this time t further vertical distance covered
would be the distance it took for final upward
velocity to become 0.
Using the equation:
2*g*s2 = v^2 - u^2
where,
g = -9.81
v = 0
u = 519.91 m/s
You will get
s2 = 13784.11 m

And,
Further horizontal distance covered d2 :
d2 = t * vcos45 [horiz. component of velocity]
where t = 137.2 s
=> d2 = 75380.694m

ANSWER
a) Maximum Altitude of rocket
= s1 + s2 = 21115.09 m + 13784.11 m = 34899.2 m
b) Rocket's total flight time
= 28s + 137.2s = 165.2 s
c) Total Range
= d1 + d2 = 23450.68 m + 75380.69 m = 98831.37 m

Hope that solves it. If not,get back to me.

Answer 3

do your own homework and quit tring to get others to answer it see im not confused i know exatly what your up to

Answer 4

You should really do your own physics homework

Answer 5

a-its up there b-its a round number c-bz3kj7ythmv..l/gfd

Answer 6

a) approx. 23336.3 m
b) approx. 147.63 sec
c) approx. 77959.4 m

Answer 7

you're right; I'm confused.

Answer 8

C) 85179.5 m
A) 21791.5 m
B) 119.327s

( i dunno if my answers are right- but i tried- please let me know the correct answers when u get them)

Answer 9

blue + red = violet
so that means ...
idk im confused