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

[negation]

Homework Statement

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

The Attempt at a Solution

t_apex = [itex]vi sin Θ/g[/itex]
t_{full trajectory} = [itex]2vi sin Θ/g[/itex]
h(t_apex) = h = [itex]vi^2 sin^2 Θ/2g[/itex]

x(t_{full trajectory}) = x = [itex]\frac{vi^2 sin (2Θ)}{g}[/itex]

 

[voko]

Consider the ratio of range to max height.

 

[negation]

Consider the ratio of range to max height.

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

 

[voko]

Show what you get and how you get that.

 

[negation]

Show what you get and how you get that.

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

 

[voko]

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

 

[negation]

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

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?

 

[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.

 

[negation]

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.

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.

 

[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?

 

[negation]

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?

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