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.(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Is the correct?

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