Why doesn't this solution work?

[dvs0826]

Why doesn't this "solution" work?

Homework Statement

A ball is to be shot from level ground toward a wall at distance x with launch angle theta. The y-component of the velocity just as it would reach the wall, as a function of that distance x is given by the equation v_y(x) = -0.5 x + 5. Find the launch angle theta.

Homework Equations

projectile motion

The Attempt at a Solution

This problem was originally shown with some graphs, but I rewrote it with an exact equation for v_y(x) instead of the graph. Hopefully it's equally clear. Henceforth I'll use v to mean v_y and a to mean a_y since I'm not interested in the x-direction for this problem.

I should say up front that I've already solved this problem "the hard way" (which was actually much harder than it appears at first glance). I had what I thought was a clever idea though and tried to apply some calculus to this problem but I seem to be doing something wrong. Essentially I said that if v = -0.5x + 5, then -0.5 = dv/dx = dv/dt * dt/dx = a(t) / v(t)

-0.5 = a(t) / v(t)

The second step comes from the definition of the chain rule, and the third from the fact that dv/dt = a(t) and dt/dx = 1/(dx/dt) = 1 / v(t). Since we know a(t) is equal to -9.8 m/s^2, and that dv/dx is -0.5 since it's a straight line, we have

-0.5 = -9.8 / v(t)
v(t) = 19.6 m/s

Obviously this is wrong since velocity in the y direction isn't going to be constant. I don't see any obvious reason why the chain rule doesn't apply here although that's the most obvious place for there to have been an error. Since I know velocity isn't constant, and that a = v', I could also have re-written the equation as:

-0.5 v = v'

This would be a differential equation with solution v(t) = Ke^(-0.5 t) + C

This doesn't seem right either though, because that means acceleration is non-constant, and this is just a simple particle in free-fall (again ignoring the x-component of motion).


So I guess I'm making a fundamental error, can anyone point it out? Thanks.

 

[rl.bhat]

vy = -0.5*x + 5.
When x = 0 Vy = 5 = Vo*sinθ...(1)
When Vy = 0, x = 10 m. = Vo*cos*θ*t = Vo*cos*θ*Vosinθ/g...(2)
From these two equations find θ.

 

[dvs0826]

vy = -0.5*x + 5.
When x = 0 Vy = 5 = Vo*sinθ...(1)
When Vy = 0, x = 10 m. = Vo*cos*θ*t = Vo*cos*θ*Vosinθ/g...(2)
From these two equations find θ.

Where are you getting t = v_0 sin(θ) / g?

Also I'm still curious why there's an error in the chain rule method above, even though this method also appears easier than the way I used to solve it.

 

[kuruman]

Your "clever" way to approach the problem is indeed a step-saver. The problem is that you did not keep track of your vector components properly when you applied the chain rule. Check this out

$$a_{y}=\frac{dv_{y}}{dt}=\frac{dv_{y}}{dx}\frac{dx}{dt}=v_{x}\frac{dv_{y}}{dx}=(-0.5 \;s^{-1})v_{x}$$

As you see, the horizontal component of the velocity is constant because the acceleration of gravity is constant. I am sure you can wrap it up from this point.

 

[rl.bhat]

Where are you getting t = v_0 sin(θ) / g?

Also I'm still curious why there's an error in the chain rule method above, even though this method also appears easier than the way I used to solve it.

When the ball reaches the highest point Vy = 0. = V0*sinθ - g*t

 

[dvs0826]

Your "clever" way to approach the problem is indeed a step-saver. The problem is that you did not keep track of your vector components properly when you applied the chain rule. Check this out

$$a_{y}=\frac{dv_{y}}{dt}=\frac{dv_{y}}{dx}\frac{dx}{dt}=v_{x}\frac{dv_{y}}{dx}=(-0.5 \;s^{-1})v_{x}$$

As you see, the horizontal component of the velocity is constant because the acceleration of gravity is constant. I am sure you can wrap it up from this point.

Doh. It always feels like such an epic fail coming up with a really cool idea and then messing up because of something simple. Based on your suggestion I get approximately 14 degrees launch angle.

 

[kuruman]

That's about what I get for the angle.

 

[dvs0826]

Edit: Nvm, thought I had a different method that wasn't working but turns out it does work, made another careless mistake.

 

[rl.bhat]

dy/dx = dy/dt*dt/dx = v_y*1/v_x = tanθ - gx/(v_o*cosθ)

so v_y = v_x*tanθ - gx/v_x
Compare it with the given equation.
You get tanθ = 5/v_x and v_x = g/0.5.

 

[dvs0826]

Yea, for some reason I forgot the chain rule needed to be applied this time. Duh :) Thanks!