Projectile motion particle question

[misogynisticfeminist]

A particle is projected from the origin 0 with initial velocity v at an angle of elevation \theta and it moves freely under gravity. Find tan \theta 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

\frac {v^2_y}{ 2g} = v_x t =v_y - 1/2 gt^2

but i am unsure of how to manipulate it from here to find v_y / v_x.

 

[marlon]

Use these

v_x = vcos( \theta)
v_y = vsin( \theta) -gt

x = vcos (\theta)t
y = vsin( \theta)t - g \frac{t^2}{2}

Then, the maximal heigth can be determined by realizing that v_y must be zero there. Calculate the time at which this occurs and put it into the formula for y. same goes for x (set y = 0 and calculate the time and put it into x) and then set these two equal to each other

marlon

 

[misogynisticfeminist]

hey thanks for the help marlon, I've gotten it...