Compute the angle θ that will maximize the height of h of the impact point

[Alexanddros81]

Homework Statement

A projectile, launched at A with an initial velovity of 24m/s at the angle θ, impacts the vertical wall at B. Compute the angle θ that will maximize the height of h of the impact point. What is this maximum height?

Homework Equations

The Attempt at a Solution

I guess I have to assume something but I can't think of what this is.
Also have a look on my previous post which is related:
https://www.physicsforums.com/threa...itial-v-24-m-s-at-angle-th-60-degrees.922032/
from previous post I can compute h in terms of θ. This is one equation with two unknowns.
So do i have to make an assumption on the velocities ?

By the way the previous post was 12.57 and this one is 12.58. If someone can correct the number of previous one mentioned in the title that would be great. thanks

 

[cnh1995]

from previous post I can compute h in terms of θ. This is one equation with two unknowns.
So do i have to make an assumption on

And you want to maximize this height by changing θ.
If h is a function of θ, how do you find θ for which h is maximum? Have you studied differentiation?

 

[kuruman]

Here is where your hunch that v_y = 0 when the projectile hits the wall comes into play. (Why?)

 

[Alexanddros81]

We assume that $v_y=0$ meaning that if there is no speed in the upward direction there is no motion upwards. The ball even if there was no wall could not get any higher for the angle calculated above.

 

[cnh1995]

View attachment 208704

We assume that $v_y=0$ meaning that if there is no speed in the upward direction there is no motion upwards. The ball even if there was no wall could not get any higher for the angle calculated above.

On the LHS, you have taken derivative of y w.r.t. time but on the LHS, you have differentiated w.r.t.θ.
It is dy/dθ=0.

Other than that, your math looks correct to me.
You could have got there faster if you'd directly differentiated sec²θ as 2sec²θtanθ. The sec²θ term can be taken out common and the rest of the LHS can be equated to 0, which directly gives you the value of tanθ.

 

[TSny]

View attachment 208704
We assume that $v_y=0$

Is the assumption $v_y=0$ correct? Note that @cnh1995 's method yields the answer without this assumption. Actually, your solution will match cnh1995's since you did set the derivative of y_{at wall} with respect to the launch angle θ equal to zero. So, you can check if your answer for θ makes $v_y=0$ at the wall.

 

[Alexanddros81]

a) I don't quite understant though (or grasp the idea of) what does the derivative of y_{at wall} with respect to the launch angle θ means (or shows)
- also when set to zero.
b) I have substitute in equation $v_y=-gt+v_0sinθ$ the angle θ with 72.98deg and i don't get v_y=0

 

[CWatters]

Perhaps worth looking at..
https://www.mathsisfun.com/calculus/maxima-minima.html
If you have an arbitrary function then the maxima and minima occur where the slope is zero. So typically you differentiate the function, equate the result to zero and find the solutions.

 

[TSny]

a) I don't quite understant though (or grasp the idea of) what does the derivative of y_{at wall} with respect to the launch angle θ means (or shows)
- also when set to zero.

As you change the launch angle, the projectile hits the wall at different heights. y_{at wall} is just notation for the height that the projectile hits the wall. It is a function of the launch angle θ. You want to maximize this function. You found an expression for this function. The value of θ that maximizes y_{at wall} is found by setting dy_{at wall}/dθ = 0 (as you essentially did).

b) I have substitute in equation $v_y=-gt+v_0sinθ$ the angle θ with 72.98deg and i don't get v_y=0

Yes. So, this shows that v_y is generally not equal to zero at the wall when the projectile hits the wall at maximum height on the wall.