Projectile Motion Challenge: Find a Difficult Problem

[abc]

difficult question !

hello all ...
i am a new member in this wonderful forum ...
i am challenging my friend in the subject ( the projectile motion) ... i wonder if i can have a very difficult problem to give him to solve
i want a very tough question ... and will be so grateful to u
thanx in advance
regards
abc

 

[Nenad]

how difficult, what grade are you in?

 

[abc]

very tough ...
i am at high school , but he is at university ...
i will be so grateful to u nenand if u give it to me as quick as possible
thanx

 

[Nenad]

the toughest one I can think of is:
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.

 

[abc]

Can U Convey The Idea By A Draw
Thanx

 

[NateTG]

Not Sadistic enough...

How about this:

The Iraqi supergun at Latitude $$x$$, longditude $$y$$ and alititude $$z$$ is going to shoot at a test target at lattitute $$x'$$ longditude $$y'$$ and altitude $$z'$$. Find the angle and velocity of the launch for minimum flight time. You may neglect air resistance, and assume that the Earth is a sphere with radius $$r_e$$ and angular velocity $$\omega_e$$.

 

[JohnDubYa]

Consider a hill with a slope of 30 degrees. At the base of the hill is a mortar (like a small cannon) that ejects shells with an initial velocity of 100 meters per second. At what angle does GI Joe have to project the mortar to hit a terrorist 50 meters up the hill?

And if he can't solve this problem, tell him the Marines have no use for him.

 

[Gza]

Good ol' president bush had one too many lines of coke to start the day, so he decides to grab Saddams gun and go a huntin' when he spots John Kerry hiding in a tree near the white house. Kerry, alarmed by his appearance abrubtly jumps out of the tree. Assuming his height to be 1.9 meters, what angle from that height above the horizontal will Dubya need to aim the gun to hit Kerry the moment he leaps, and secure his spot in the white house for another four years?

 

[abc]

what tough questions
thanx all ! i feel dizzy after reading them
i am sure he wouldn't be able to solute them
but he'll at least try

 

[Nenad]

lol, that last one is funny.

 

[NateTG]

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.

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.

 

[Rogerio]

the toughest one I can think of is:
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.

Since the angles are the same (30 degrees), points P and O are at the same level.

Considering 'O' we can say that .5mv^2 = sqrt(mgh) or v=sqrt(2gh)
The angle is 30degrees, so vx= v*cos(30) and vy=v*sin(30)

It takes time t to revert the initial vertical velocity from vy to -vy , at point P, or:
gt= 2vy , which gives t=2vy/g = v/g

Then, the distance OP is
vx*t = v * cos(30) * (v/g) = sqrt(2gh) * sqrt(3)/2 * sqrt(2gh) / g

OP = h*sqrt(3)/2 = 15.6 m

[]s

-------
Editing:
I've considered the skier lands at point P with 30 degrees below the horizontal.
But it seems the 30 degrees is between OP and the horizontal. In this case, the NateTG's response is the correct one.

 

[Nenad]

NateTG is right. Good job.

 

[pervect]

hello all ...
i am a new member in this wonderful forum ...
i am challenging my friend in the subject ( the projectile motion) ... i wonder if i can have a very difficult problem to give him to solve
i want a very tough question ... and will be so grateful to u
thanx in advance
regards
abc

Just ask him to write down the equations of motion for a ball with a moment of inertia I, a coefficient of static friction m_s, and a coefficient of rolling fricition m_r moving n a hemispherical bowl...

 

[Link]

how do you get the equation images you can click on?

 

[NateTG]

how do you get the equation images you can click on?

There is tex support on this site. Look for the "Introducing LaTeX typestting" thread for more info.

 

[SergejVictorov]

Ok, I've got a VERY difficult one for you guys!

All units are metric. This is projectile motion WITH air resistance!

The muzzle of a cannon is placed at the point of origin in a 3d grid (P(0;0;0)). It fires a spherical aluminium projectile with a radius of 3cm at a velocity of 150 m/s.
A constant wind is blowing in the positive y-direction with 5 m/s.

1) A circlular target area with a radius of 10m lies on the Earth's surface (the x-y plane). Its center is at (150;25;0). This area must be hit. Find the approximate range of angles between the cannon and the x-y plane (elevation) as well as the angle between the cannon and the x-z plane.

2) Where would the projectile land when fired at the same velocity as before with an elevation of 23 degrees and in the direction of the x-axis? What would its impact velocity be?

If you ever feel bored, try to solve it. Actually, it is very easy!

 

[Link]

LOL its hard for a high school student