Projectile Motion Problem with thrown ball off roof

[coco87]

Homework Statement

A ball is thrown leftward from the left edge of the roof, at height [tex]h[/tex] above the ground. The ball hits

the ground [tex]1.50s[/tex] later, at distance [tex]d=25.0m[/tex] from the building and at angle [tex]\theta =60^

{\circ}[/tex] with the horizontal. (a) Find h. What are the (b) Magnitude and (c) angle relative to the horizontal

of the velocity at which the ball is thrown? (d) Is the angle above or below the horizontal?

http://lcphr3ak.is-a-geek.com/fig49.png

[tex]d=25.0m[/tex]
[tex]\theta=60^{\circ}[/tex]

The answers in the back are:
a)[tex]32.3m[/tex]
b)[tex]21.9\frac{m}{s}[/tex]
c)[tex]40.4^{\circ}[/tex]
d)below

Homework Equations

[tex]x - x_{\circ} = v_{\circ} t - \frac{1}{2} a t^2[/tex]
[tex]v_{\circ x} = v_{\circ} \cos{\theta _{\circ}}[/tex]
[tex]v_{\circ y} = v_{\circ} \sin{\theta _{\circ}}[/tex]

The Attempt at a Solution

I'm missing an observation here, because I don't see enough information to solve this. This problem is suppose to

be solved with projectile motion. So first I attempt to find [tex]v_{fx}[/tex] (which is the [tex]x[/tex] component

of the vector at [tex]\theta[/tex]): [tex]x - x_{\circ} = v_{fx} t - \frac{1}{2} a t^2[/tex] which is [tex]-25 =

v_{fx} (1.50)[/tex]. This gives me [tex]v_{fx} = -16.667[/tex]. Using this, I take [tex]v_{fx} = v_f \cos{\theta}

[/tex], which is: [tex]-16.667 = v_f \cos{60}[/tex], so [tex]v_f = -33.32[/tex]. Using this, I find [tex]v_{fy}

[/tex] by taking [tex]v_{fy} = v_f \sin{\theta}[/tex] and doing [tex]v_{fy} = (-33.32)\sin{60}[/tex] so [tex]v_{fy}

= -28.856[/tex]. So, to find [tex]h[/tex], we solve for [tex]y[/tex]. [tex]y = v_{fy} t - \frac{1}{2} a t^2[/tex]

becomes [tex]y = (-28.856)(1.50) - \frac{1}{2} (9.8) (1.50)^2[/tex] which says [tex]y = -54.309[/tex]. Now, the

absolute value of [tex]y[/tex] should be the height, but this is waaaaay off from the answer. I'm practically out of

ideas.

Could anyone explain what I'm doing wrong? If I didn't provide enough info, just let me know and I'll try to

provide more.

Thanks!

 

[tiny-tim]

Hi coco87!

(have a theta: θ )

… Using this, I take [tex]v_{fx} = v_f \cos{\theta}

[/tex], which is: [tex]-16.667 = v_f \cos{60}[/tex], so [tex]v_f = -33.32[/tex]. Using this, I find [tex]v_{fy}

[/tex] by taking [tex]v_{fy} = v_f \sin{\theta}[/tex] and doing [tex]v_{fy} = (-33.32)\sin{60}[/tex] so [tex]v_{fy}

= -28.856[/tex].…

Nooo … θ in your formula is the angle at the roof. not the θ (= 60º) in the diagram!

 

[disillusion]

Take care with the sign of the acceleration.
It is often helpful to write down on your diagram which direction you are taking as positive.

 

[coco87]

Hi coco87!

(have a theta: θ )


Nooo … θ in your formula is the angle at the roof. not the θ (= 60º) in the diagram!

So, I'm either using the wrong formula, or I need to find the initial velocity (or angle)? I don't know of any other formulas (other than [tex]v = v_{\circ}+at[/tex]), and I have no idea how I would even obtain the angle up there. Could you possibly give me a hint? (or another one )

Thanks for the responce!

 

[disillusion]

So, I'm either using the wrong formula, or I need to find the initial velocity (or angle)? I don't know of any other formulas (other than [tex]v = v_{\circ}+at[/tex]), and I have no idea how I would even obtain the angle up there. Could you possibly give me a hint? (or another one )

Thanks for the responce!

you can use that formula to find the initial vertical velocity. Then you can work out an angle!

 

[coco87]

you can use that formula to find the initial vertical velocity. Then you can work out an angle!

What you had said triggered an idea, and after playing around with the equations, I found out that I had misunderstood the Constant Accelloration equations all along.. well, I do now

Thank you guys for your help!