Nenad said:
lol, that last one is funny.
Especially since there isn't enough information to solve it :). Of course, it's not that hard since Bush just has to aim directly at kerry.
Nenead said:
A skier goes down a parabolic slope of a hill.(the slope goes straight down then up is a parabolic arc). The point at which he flies off the hill is called Point O, and the vertical height the skier drops befoure hitting point O is 18m. His takeoff angle is 30 degrees above the horizontal, and he eventually lands at point P which is 30 degrees below the horizontal. What is the length of OP is metres. Neglect all friction and air resistance, and treat the skier at a point mass.
Isn't exactly clear either. Do you mean:
A skier skis frictionlessly from a standing start, loosing 18m of altitude before hitting a jump that lauches him at an angle of 30 degrees above the horizontal. The slope from the launch point is 30 degrees below the horizontal. Find the distance (along the slope) between the jump and the point where the skier lands.
This isn't so bad:
The skier's lanunch velocity can be found using energy:
\frac{1}{2}m v^2=mgh
|v| = \sqrt{2gh}
Then it's possible to find the x and y components of the lauch velocity:
v_{x}= \cos{\theta}|v|
and
v_{y}=\sin{\theta} |v|-gt
so
p_x(t)=t |v| \cos{\theta}
p_y(t)=t |v| \sin{\theta} - \frac{1}{2} gt^2
where \theta is the launch angle.
And the point where the skier hits is where
\frac{p_y}{p_x}=\tan{\phi}
Solving for t yields:
t=\frac{2v}{g}(\sin \theta - \frac{\sin \phi \cos \theta}{\cos \phi})
Now, this can be plugged back into the formulas to get the distance:
\frac{2v^2}{g}\frac{\cos \theta}{\cos \phi} (\sin \theta - \frac{\sin\phi \cos\theta}{\cos{\phi}})
or
4h \frac{\cos \theta}{\cos \phi} (\sin \theta - \frac{\sin\phi \cos\theta}{\cos{\phi}})
Plugging in your numbers gives a total distance of 72m.
