Is the correct?

  • Thread starter gimpy
  • Start date
  • #1
28
0

Main Question or Discussion Point

Ok i just want to comfirm that i did this correctly. I am making a simulator for a missile launch (simple java program). Anyways the Launch pad is located at 0m and the target is located 250m away and it is 170m high and flat and infinitly long. Now i want the missile to just display where it hits on the target (170m high, if it can fit on the screen or course). So this is what i did.

I used for the x and y positions

x = t*V*cos(alpha)
y = t*V*sin(alpha) - (g*t^2)/2

t = ?
V = 200 m/s
g = 9.81 m/s^2
alpha = 45 degrees * PI/180 (conversion to radians)

Now i know what y is (170m) so to find x (impact point) i must solve the equation y = t*V*sin(alpha) - (g*t^2)/2 for t then substitute t into x = t*V*cos(alpha).

So,

(gt^2)/2 - t*V*sin(alpha) + y = 0

then i solve this using the quadratic equation

(-b +- sqrt(b^2 - 4*a*c))/2*a

where
a = g/2
b = -(V*sin(alpha))
c = y

to obtain the roots , which are the times in seconds where the missile is at h=170m (going up and coming down). So obviously the the greater root (t) is the impact point of t (where the projectile lands on the target when it comes down).

Oh and to solve the quadratic equation i divided a, b and c by a, just to make a = 1. ( i get a different answer if i don't )

Does this sound right? Im pretty sure it is, i just want to make sure.

Thanks
 

Answers and Replies

  • #2
Integral
Staff Emeritus
Science Advisor
Gold Member
7,199
56
What you appear to be modeling is the case where a body is moving only under the influence of gravity. Is this correct for a rocket? Does it not start at zero velocity, accelerate to a maximum velocity at some altitude then begin the free fall situation? You need to investigate the rocket equation which takes into consideration the changing mass as the fuel burns during the acceleration phase.

What you are modeling is simply a rock thrown up at an angle with some velocity, this is not a rocket.
 

Related Threads on Is the correct?

Replies
9
Views
2K
  • Last Post
Replies
4
Views
1K
Replies
22
Views
4K
Replies
7
Views
2K
Replies
0
Views
350
  • Last Post
Replies
3
Views
588
Replies
5
Views
2K
  • Last Post
Replies
11
Views
685
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
9
Views
1K
Top