- #1
misogynisticfeminist
- 370
- 0
A particle is projected from the origin 0 with initial velocity v at an angle of elevation [tex] \theta [/tex] and it moves freely under gravity. Find [tex] tan \theta [/tex] if the greatest height reached by the particle is equals to its range on the horizontal plane.
Using the equations of motion, i have found
[tex] \frac {v^2_y}{ 2g} = v_x t =v_y - 1/2 gt^2 [/tex]
but i am unsure of how to manipulate it from here to find [tex] v_y / v_x [/tex].
Using the equations of motion, i have found
[tex] \frac {v^2_y}{ 2g} = v_x t =v_y - 1/2 gt^2 [/tex]
but i am unsure of how to manipulate it from here to find [tex] v_y / v_x [/tex].