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What is the Maximum Height of this projectile?

1. Homework Statement
A projectile is launched at 750 meters per second at a 30 degree angle to the horizon. How long is it in the air? How far does it go? What is the maximum height that it reaches?

2. Homework Equations
Sin(30)=1/2
Cos(30)=sqrt(3)/2

3. The Attempt at a Solution
This is what I have so far. I believe I successfully solved for the time it was in the air, as well as how far it traveled (I used the method from Khan Academy, rather than my professor's method, since all he gave us were a bunch of equations and didn't explain which one gave what), and I came out with these answers. It would be super helpful if I could get confirmation on whether I'm correct or not. Now, the thing I'm stuck at is how high the projectile must be at its highest. I assume this means that it needs to be at the halfway point of the total displacement, but I don't know where to go from there.

W5nAFOd.jpg
 

Student100

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1. Homework Statement
A projectile is launched at 750 meters per second at a 30 degree angle to the horizon. How long is it in the air? How far does it go? What is the maximum height that it reaches?

2. Homework Equations
Sin(30)=1/2
Cos(30)=sqrt(3)/2

3. The Attempt at a Solution
This is what I have so far. I believe I successfully solved for the time it was in the air, as well as how far it traveled (I used the method from Khan Academy, rather than my professor's method, since all he gave us were a bunch of equations and didn't explain which one gave what), and I came out with these answers. It would be super helpful if I could get confirmation on whether I'm correct or not. Now, the thing I'm stuck at is how high the projectile must be at its highest. I assume this means that it needs to be at the halfway point of the total displacement, but I don't know where to go from there.

W5nAFOd.jpg
I didn't explicitly check your numbers, but it looks okay.

Also, for future problems, the only thing you need to remember here is that ##F=ma## and then examine the specifics of the case. Where ##a = \frac{d^2x}{dt^2}##

In the x direction, acceleration is equal to 0, which also suggests that ##F_x## is equal to 0. So you're left with ##0=m \frac{d^2x}{dt}## Integrating a few times will give you ##v_x = v_{x0}## and ## x_x = x_{x0} + v_{x0}t## which describes it's motion completely.

In the y direction, acceleration is a constant g, which then suggests ##F_y## is equal to some constant, which happens to be mg. So ##F_y=m \frac{d^2x}{dt^2}## or ##mg = m \frac{d^2x}{dt^2}## simplifying and doing the calculus leaves you with ##v_y = v_{y0} + gt## and ##x_y = x_{y0} + v_{y0}t +\frac{1}{2}gt^2##

To arrive at something like the max height equation you're simply combining equations... ##v_y^2 = v_0^2 + 2g(x_y-x_{y0})## which itself comes from a combination of the above and reasoning it out. ##x_{y0}## is your height at launch. What is the velocity at the maximum height in the y direction? Can you derive this?

Then simply solve the equation for ##x_y##

I left some work here for you, hopefully you see the importance of thinking in physics, from a three letter equation you can derive everything you need. :smile: Also, solve everything as variables before you plug in, it makes your work easier to read and you'll be less mistake prone.
 

haruspex

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Your answers are basically correct. You appear to have used 10ms-2 for g, so you should not quote any greater precision for the answers. E.g. the distance should be 50km.
Are you sure you should be using 10, rather than, say, 9.8?
 

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