physics - angle for particular co-ordinate in projectile path? -


I have xy co-ordination (200,200). I know that the ball throws the angle from the original, the ball is thrown into the 2D environment, so how can I get the initial speed to reach that specific XY co-ordination?

Use of Iam

  x = v0cosq0t; Y = v0sinq0t - (1/2) gt2  

But does the time need to be done without time? Any help please?

I'm assuming that you get the ball at that particular point (200,200) at the top of your path, My physical science is a bit confusing, but it is that I have thrown together:

v_y = square_root (2 * g * y) ,

where G is a serious number, which shows acceleration due to gravity, and y how high you want to go (200 in this case).

v_x = (x * g) / v_y ,

Where is the distance in the X-X (200 in this case), like the previous one , And Vy's answer we got in the previous equation.

This equation eliminates the need for an angle However, if you have velocity + angles, then it is easy:

v0 = square_root (v_x ^ 2 + v_y ^ 2)

and

angle = arctan (v_y / v_x) .

Here's the etymology, if you are interested:

(1 / 2) ^ 2 + v_yt - y = 0

t = (-v_y +/- square_root (v_y ^ 2 - 2 We also have another equation, because the vertical velocity at the top is 0:

0 = v_y + at / P>

Options:

0 = v_y + (-v_y +/- square_root (v_y ^ 2 - 2))

0 = square_root (v_y ^ 2 - 2))

0 = v_y ^ 2 - 2 app

v_y = For square_root (-2y) , or

v_y = square_root (2gy)

v_x:

< Before code> v_x * t = x

, t = v_y / a, such

v_x = (x * G) / v_y

I hope enough make sense.


Comments

Popular posts from this blog

sql - dynamically varied number of conditions in the 'where' statement using LINQ -

asp.net mvc - Dynamically Generated Ajax.BeginForm -

Debug on symbian -