Homework Help: Show that the horizontal range is 4h/tan(theta)

1. Jan 27, 2014

negation

1. The problem statement, all variables and given/known data

A projectile launched at angle Θ to the horizontal reaches maximum height h. Show that its horizontal range is 4h/ tan Θ.

3. The attempt at a solution

tapex = $vi sin Θ/g$
tfull trajectory = $2vi sin Θ/g$
h(tapex) = h = $vi^2 sin^2 Θ/2g$

x(tfull trajectory) = x = $\frac{vi^2 sin (2Θ)}{g}$

2. Jan 27, 2014

voko

Consider the ratio of range to max height.

3. Jan 27, 2014

negation

I did. The expression for the full range and heigh, h is in the op.
But I'm unable to reduce them to the appropriate wxpression

4. Jan 27, 2014

voko

Show what you get and how you get that.

5. Jan 27, 2014

negation

I worked out hapex = vi ^2 sin^Θ and xfull range = vi^2(2Θ)/g
The above was obtained by substituting t/2 intp the y-displacement and t into the x-displacement.

6. Jan 27, 2014

voko

As I said. Consider the ratio of range to max height.

7. Jan 27, 2014

negation

I did.

The horizontal range is 2vi^2 sinΘcosΘ.
I simplified 4h/tanΘ to 2vi^2sinΘcosΘ/g.
I suppose this is a sufficient condition for the proof?

8. Jan 28, 2014

voko

I cannot see your analysis of the ratio of range to max height. Which is strange, because you have found a formula for range, and a formula for max height. All you need is to divide one by another.

9. Jan 28, 2014

negation

I failed to obtain the required equation. Why do we need the ratio? The question has not asked for that.

My method:
I found x = vi^2 sin(2Θ)/g = 2vi^2 sinΘcosΘ
I reduced 4h/tanΘ to = 2vi^2 sinΘcosΘ, where h = vi^2 sin^2Θ/2g and tan Θ = sinΘ/cosΘ.

10. Jan 28, 2014

voko

You are required to prove that $$x = {4h \over \tan \theta}$$ That means $${x \over h} = {4 \over \tan \theta}$$

Regarding your "failure", how are $\cot$ and $\tan$ related?

11. Jan 28, 2014

negation

cot Θ= 1/tanΘ
It never occurred to me I had to perform a x/h ratio. But anyway, answer found.

Last edited: Jan 28, 2014