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

• Benjamin_harsh
In summary: Good job!In summary, the time it takes for an object to travel from point A to point B can be calculated using the equation ##T = \frac{V\sin\theta}{g}##. The maximum height that the object can reach is given by the equation ##h = \frac{V^2\sin^2\theta}{2g}##, and the time it takes for an object to travel from point B to point C can be found using the formula ##T = \sqrt{\frac{2y_{max}}{g}}##. The proper formula for ##y_{max}## for this trajectory is ##y_{max} = \frac{V^2\sin^2\theta}{
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?

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

Last edited:
Benjamin_harsh said:
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.

Horizontal range is ##R = Vcosθ.t##

Benjamin_harsh said:
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?

Benjamin_harsh said:
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.

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

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

##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}##

jbriggs444
Benjamin_harsh said:
##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.

## 1. What is the definition of "Ymax" in the context of a trajectory?

Ymax is the maximum vertical distance reached by an object in a given trajectory. It is typically measured from the initial starting point or ground level.

## 2. How do you calculate the value of Ymax for a given trajectory?

The formula for calculating Ymax depends on the specific trajectory and the variables involved. Generally, it involves solving for the maximum height in the equation of motion for the vertical direction.

## 3. Is there a standard formula for calculating Ymax for all trajectories?

No, there is not a single formula that can be applied to all trajectories. The formula for Ymax will vary depending on the specific variables and conditions of the trajectory.

## 4. Can Ymax be negative in a trajectory?

Yes, Ymax can be negative if the initial starting point or ground level is taken as the reference point. In this case, Ymax would represent the lowest point reached by the object in its trajectory.

## 5. How can Ymax be affected by external factors such as air resistance or gravity?

External factors such as air resistance and gravity can affect the value of Ymax by altering the trajectory of the object. Air resistance can decrease the maximum height reached, while gravity can increase or decrease it depending on the direction of the force.

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