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garroad_ran
Registered User regular

I never had a chance to take a physics course in high school and I've been trying to get into it over the last couple of weeks.

I've been at this problem for a couple of days, but I can't seem to break through. I suppose I could just look at the answer and explanation in the back of the book, but if anyone can give me an idea of how I'm supposed to progress here that would be far preferable to having the answer given to me.

The problem:

A small projectile is launched from the floor at an angle toward a wall.

The distance from the wall is l.

The initial speed is v0.

The angle is theta.

The only force we are considering is gravity.

Find the time it takes for the projectile to reach the wall (t1) and the height (h) where it will impact the wall.

Here's what I've got so far.

Now I have to find the height of the point of impact (h).

It feels like it should be easy enough to do, using this equation:

h = (v)(t) + (1/2)(g)(t^2)

But I can't seem to come up with anything nicer than

h = (sin(theta))(l) / (cos(theta)) + (g)(l^2) / 2(v0^2)(cos(theta)^2)

I can't see that I've made any algebraic mistakes, but I also can't believe that that's the equation this book is expecting me to come up with...

I've been at this problem for a couple of days, but I can't seem to break through. I suppose I could just look at the answer and explanation in the back of the book, but if anyone can give me an idea of how I'm supposed to progress here that would be far preferable to having the answer given to me.

The problem:

A small projectile is launched from the floor at an angle toward a wall.

The distance from the wall is l.

The initial speed is v0.

The angle is theta.

The only force we are considering is gravity.

Find the time it takes for the projectile to reach the wall (t1) and the height (h) where it will impact the wall.

Here's what I've got so far.

Start by separating the x and y components of velocity. Draw a right triangle with vertices A(launch point) B(impact point) C(point where the wall meets the ground).

sin(theta) = opp/hyp = initial vertical velocity / v0

cos(theta) = adj/hyp = initial horizontal velocity / v0

I can rewrite those such that:

initial vertical velocity = (v0)(sin(theta))

initial horizontal velocity = (v0)(cos(theta))

With that, figuring out the time to impact is a piece of cake.

l = (initial horizontal velocity) (t1)

t1 = l / (initial horizontal velocity)

or

t1= l / (v0)(cos(theta))

sin(theta) = opp/hyp = initial vertical velocity / v0

cos(theta) = adj/hyp = initial horizontal velocity / v0

I can rewrite those such that:

initial vertical velocity = (v0)(sin(theta))

initial horizontal velocity = (v0)(cos(theta))

With that, figuring out the time to impact is a piece of cake.

l = (initial horizontal velocity) (t1)

t1 = l / (initial horizontal velocity)

or

t1= l / (v0)(cos(theta))

Now I have to find the height of the point of impact (h).

It feels like it should be easy enough to do, using this equation:

h = (v)(t) + (1/2)(g)(t^2)

But I can't seem to come up with anything nicer than

h = (sin(theta))(l) / (cos(theta)) + (g)(l^2) / 2(v0^2)(cos(theta)^2)

I can't see that I've made any algebraic mistakes, but I also can't believe that that's the equation this book is expecting me to come up with...

0

## Posts

The gravitational equation is second order, and I don't see an easy solution to clean it up. I'd probably keep all the quadratic parts within brackets, as g(l/(v0.cos(theta)))^2. Typed out in text it looks a bit ugly, but on paper or with decent notational software it doesn't look too complex. There are transformations that you could do to lose the inverted quadratic cosine (cos x)^2 = cos(2x)/2 +1/2 and 1/cos(x)=arccos(x) but it'd balloon the equation.

Also note that it depends a bit on the textbook, but in that formula the gravitational constant has to be defined as negative (because the movement is of course downward).

edit: Doesn't matter anyway. Your arithmetic isn't wrong, you've just missed something that can make the function a lot cleaner looking. You can reduce the height equation to only contain tan and tan^2 as trigonometric terms. Or tan and sec^2 if you like the way that looks better. Look back at your trig identities.

tarnokon0431-6094-6446-7088

I followed through to the end of the problem, and except for a small error in reasoning on the very last question, everything checked out.