# Help w/ projecetile motion w/ angled launch problem

• physicsgeek54
In summary: I had to do the same thing for the horizontal and it was really confusing! Once I solved for the horizontal and vertical components separately everything made a lot more sense.

## Homework Statement

A bowling ball rolls off an angled roof. The roof is sloped downward at 18.4°, and the eaves are 12.0m above the ground.If it lands 2.0m away from the building how fast was it rolling when it left the roof(i.e find Vi).

## Homework Equations

I know I have to use the equation Vh=Δx/Δt. I then solved for Vi(which took me quite a while considering that Vi was on both sides of the equation and then I had to isolate it).

## The Attempt at a Solution

This is the most complex problem I've ever seen as Vi was on both sides(was told by teacher to isolate radical to find Vi). I was eventually able to isolate Vi on one side and got the equation:
Vi=[-(2ghi+((g^2)(Δx^2))/((cos^2θ)(sin^3θ))]^1/6
This equation looks really funky so I'm not sure if its right, I got an estimated value answer of 5m/s.

P.S For clarification on the equation, the g and Δx are both divided by the sin and cos.

I can't make heads nor tails out of your formula. This projectile problem should not lead to anything with sines cubed and sixth roots of things.

How did you begin? Did you write your kinematic equations of motion for the horizontal and vertical components of the velocity? If you did, what are they?

Well I started w/ the equation Vi cosθ=Δx/Δt and found that Vi= Δx/Δt*cosθ. Then I found Δt by using the equation Δx=(Vi)v*Δt+.5(Δt)^2 and got Δt=-(Vi sinθ±√Vi^2sin^2θ-2ghi)/g(everything from Vi^2 to 2ghi is under radical). Then I plugged in Δt and got:

Vi=(Δx*g)/(cosθ(-Vi sinθ±√Vi^2sinθ-2ghi))
Since Vi is on both sides I had to isolate the radical and that's why I came up with the equation
Vi=[-(2ghi+((g^2)(Δx^2))/((cos^2θ)(sin^3θ))]^1/6

physicsgeek54 said:
Well I started w/ the equation Vi cosθ=Δx/Δt and found that Vi= Δx/Δt*cosθ. Then I found Δt by using the equation Δx=(Vi)v*Δt+.5(Δt)^2 and got Δt=-(Vi sinθ±√Vi^2sin^2θ-2ghi)/g(everything from Vi^2 to 2ghi is under radical). Then I plugged in Δt and got:

Vi=(Δx*g)/(cosθ(-Vi sinθ±√Vi^2sinθ-2ghi))
Since Vi is on both sides I had to isolate the radical and that's why I came up with the equation
Vi=[-(2ghi+((g^2)(Δx^2))/((cos^2θ)(sin^3θ))]^1/6

It looks like you've got a mish-mash of horizontal and vertical motions wrapped up in your equations, which are not in standard form. That's going to cause hardships. You need to identify the separate motions at the outset.

Suppose that the initial speed of the projectile is v. Then the initial horizontal and vertical components of v can be written as:

vx = v*cos(θ)
vy = v*sin(θ)

The equations of motion for the separate components are then:

x = v*cos(θ)*t
y = yo + v*sin(θ)*t - (1/2)*g*t2

You can identify the known values in the above from the problem statement, namely that yo = 12.0m and θ = -18.4°, and that when the ball lands, x = 2.0m.

I suggest that you solve the equation for the horizontal motion for t and plug the result into the equation for vertical motion. You should end up with a single v2 term to isolate.

Oh wow, that really helps thanks! I made this problem a whole lot more complex than it was because when I solved for Δt I solved for the vertical component for some reason...