Projectile motion particle question

1. Apr 7, 2005

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 $$v_y / v_x$$.

2. Apr 7, 2005

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

Last edited: Apr 7, 2005
3. Apr 9, 2005

misogynisticfeminist

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