What is the proper formula for ##Y_{max}## for this trajectory?

[Benjamin_harsh]

Homework Statement

What is the proper formula for $Y_{max}$ for this trajectory?

Relevant Equations

What is the proper formula for $Y_{max}$ for this trajectory?

Time it takes from A to travel to B; $T = \large\frac{Vsinθ}{g}$

Max height object can reach, $h = \large\frac{V^{2} sin⁡2θ}{g}$

Time it takes for for B to travel to C; $T = \large\sqrt\frac{2y_{max}}{g}$

What is the proper formula for $Y_{max}$ for this trajectory?

 

[jbriggs444]

[...]
Max height object can reach, $h = \large\frac{V^{2} sin⁡^2θ}{g}$
[...]
What is the proper formula for $Y_{max}$ for this trajectory?

[Corrected your LaTeX]
You can read the answer from the givens on the drawing and from one other number you have already calculated.

Edit: Note that @kuruman interpreted your error differently (and likely more correctly).

 

[kuruman]

Max height object can reach, $h = \large\frac{V^{2} sin⁡2θ}{g}$

This is incorrect. It represents the horizontal range, not the maximum height, of a projectile when it returns to the same level from which it was launched.

 

[Benjamin_harsh]

Horizontal range is $R = Vcosθ.t$

 

[jbriggs444]

Horizontal range is $R = Vcosθ.t$

Yes, and if you set define t as the time taken to return to the starting level you get $t=\frac{2V\sin\theta}{g}$
If you then use the trig identity that $\sin 2\theta = 2\sin \theta \cos \theta$ then you get $R=\frac{V^2\sin 2\theta}{g}$

Did you mean to write $\sin 2\theta$ or $\sin^2\theta$ in your original post?

 

[kuruman]

Horizontal range is $R = Vcosθ.t$

Really? What is the range at $t=0$? The equation you quoted is the horizontal position at any time $t$ from $t=0$ to the time it lands.

 

[Benjamin_harsh]

Then what is the formula for Max height & $Y_{max}$?

 

[jbriggs444]

You tell us. Given the initial vertical velocity of a projectile, how high can it get?

 

[Benjamin_harsh]

$y_{max} = \large\frac{V^{2}sin^{2}θ}{2g}\normalsize + H$ (Here $H$ is the height of the cliff.)

$h = \large\frac{V^{2}sin^{2}θ}{2g}$

 

[kuruman]

$y_{max} = \large\frac{V^{2}sin^{2}θ}{2g}\normalsize + H$ (Here $H$ is the height of the cliff.)

$h = \large\frac{V^{2}sin^{2}θ}{2g}$

That is correct.