- #1
coco87
- 14
- 0
Homework Statement
A ball is thrown leftward from the left edge of the roof, at height h above the ground. The ball hits
the ground 1.50s later, at distance d=25.0m from the building and at angle \theta =60^<br /> <br /> {\circ} 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
d=25.0m
\theta=60^{\circ}
The answers in the back are:
a)32.3m
b)21.9\frac{m}{s}
c)40.4^{\circ}
d)below
Homework Equations
x - x_{\circ} = v_{\circ} t - \frac{1}{2} a t^2
v_{\circ x} = v_{\circ} \cos{\theta _{\circ}}
v_{\circ y} = v_{\circ} \sin{\theta _{\circ}}
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 v_{fx} (which is the x component
of the vector at \theta): x - x_{\circ} = v_{fx} t - \frac{1}{2} a t^2 which is -25 = <br /> <br /> v_{fx} (1.50). This gives me v_{fx} = -16.667. Using this, I take v_{fx} = v_f \cos{\theta}<br /> <br />, which is: -16.667 = v_f \cos{60}, so v_f = -33.32. Using this, I find v_{fy}<br /> <br /> by taking v_{fy} = v_f \sin{\theta} and doing v_{fy} = (-33.32)\sin{60} so v_{fy} <br /> <br /> = -28.856. So, to find h, we solve for y. y = v_{fy} t - \frac{1}{2} a t^2
becomes y = (-28.856)(1.50) - \frac{1}{2} (9.8) (1.50)^2 which says y = -54.309. Now, the
absolute value of y 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!
Last edited by a moderator:
)
