Projectile Motion word problem from Calc III course (finding height of a wall).

[jtex316]

The problem is from the MIT Open CourseWare Calculus eBook by Strang. Chapter 12.2, #3:

A ball is thrown at 60 degrees with initial velocity of 20m/sec to clear a wall 2m high. How far away is the wall?

Finding flight time T, range R, and max height Y(max) is the easy part. How do I find how far away the 2m-high wall is? I understand that this has two solutions (the wall is going to be really close to the launch point, or really close to the end point). How do I calculate the solutions? Please note that it's been a very long time since Physics I.

I've attempted to take the Range and subtract the height of the wall from it, but that doesn't work. I feel like instead of Y(max) I need Y(2m), but I don't know how to do that.

You don't need to work out the problem - a link to an example or a little push is all I'm looking for here.

Thanks!

 

[vela]

The problem is from the MIT Open CourseWare Calculus eBook by Strang. Chapter 12.2, #3:

A ball is thrown at 60 degrees with initial velocity of 20m/sec to clear a wall 2m high. How far away is the wall?

Finding flight time T, range R, and max height Y(max) is the easy part. How do I find how far away the 2m-high wall is? I understand that this has two solutions (the wall is going to be really close to the launch point, or really close to the end point). How do I calculate the solutions? Please note that it's been a very long time since Physics I.

I've attempted to take the Range and subtract the height of the wall from it, but that doesn't work.

The range is the distance the projectile moves horizontally, right? It doesn't really make sense to subtract a vertical distance from it. Remember that the motion in the horizontal and vertical directions are separate from each other.

I feel like instead of Y(max) I need Y(2m), but I don't know how to do that.

I'm not clear on your notation here. Y_max usually denotes the maximum height the projectile reaches. Y_{2 m} would tautologically be equal to 2 m.

You don't need to work out the problem - a link to an example or a little push is all I'm looking for here.

Start by finding the time it takes for the ball to reach a height of 2 meters.

 

[jtex316]

Hey there, thanks for your reply. Apparently it looks like I'm going to need to see the problem worked out. I have struggled with this, trying to use T, R, and Y(max) formulas and just don't see how to get to the answers as the author has.

 

[HallsofIvy]

The only way you can do this problem is to assume they mean that the ball just barely clears the wall. That is, as vela said, solve for the time it takes the ball to reach a height of 2m. Use that height to determine the horizontal distance the ball has gone.

If you have tried this yourself but don't get the answers the author has, show us what you have done, what answer you got, and what the author's answer is.