How Does Lowering Density Affect Trajectory?

[jillz]

why does lowering the density change the trajectory?

why does lowering the density change the trajectory?
Is this a basic/general question or is more information needed??
totally lost...

 

[DyslexicHobo]

Not sure what you're asking here...

If you're asking about an object that is put into flight, and then has mass taken away while maintaining the same volume, then it can easily be seen that the trajectory will change by looking at its kinetic energy. If the mass goes down, the velocity has to go up, thus changing the trajectory from its original path.

Not sure exactly when something like that could happen in real life, though.

 

[tiny-tim]

why does lowering the density change the trajectory?
Is this a basic/general question or is more information needed??
totally lost...

Hi jillz! Welcome to PF!

If you're talkling about light, it's because a higher density slows the light down (as one would expect). For details, see:

http://en.wikipedia.org/wiki/Refraction​

 

[Danger]

Echo that welcome, Jillz.
In simple terms, anything traveling through a medium requires a certain amount of energy to do so, as it has to push stuff out of the way in order to progress. The denser that medium is, the more energy is needed because it's harder to push. (Think of swinging your arm around under water as opposed to in air.) Something in ballistic flight (ie: unpowered after launch) will run out of momentum sooner and thus start back down closer to the launch site.
Even without gravity, the trajectory will change in that it will take longer to get to where it's headed or even stop completely before getting there, even though the actual flight path doesn't change.

 

[jillz]

With all of the calculations/data below, is it still possible to calculate the dimensions for the motorcycle falling/landing from the jump? Or is more data needed? What type of calculation would be used to find the dimensions for a falling object?? Thanks for any help! Jillz


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Stunt man Joe wants to cross the Grand Canyon on his motorcycle. His crew has set up a ramp on one side of the canyon. Several factors will determine his safe landing on the other side. Assume that he decides to perform the stunt on a quiet day so that the drag from the wind will not have any effect on his motion.

Assume that Joe has to cross a distance of 40 m, leaving some clearance on both sides. The clearance on either side will be 6 m. Total horizontal distance that he needs to cover is 52 m to safely land on the other side.

The crew decides to set up the ramp at an inclination of 45o.
i.e. the slope of the ramp, θ = 45o

His motion from the time he leaves the ramp to the point where he is at the same level on the other side of the canyon. During the time he is airborne, his acceleration is the acceleration of gravity.

Using s = ut + ½ a t2
s = arc length

Horizontally: 52 = u cos θ t

t = 52 / u cos θ ------------------------------- (1)

Net vertical distance he traveled = 0

Vertically: 0 = u sin θ t - ½ g t2

u sin θ = ½ g t

Substituting from (1)

u sin θ = ½ g (52 / u cos θ)
u2 = 52g/ 2 cos θ sin θ = 52g/ 2 cos 45 sin 45 = 509.6

u (speed) = 22.6 m/s

He should reach a speed of 22.6 m/s (49.7 mi/h) at the end of the ramp in order to safely cross the Grand Canyon.

 

[jillz]

thanks for the input! and cute pict!

thanks for the input! and cute pict!

Echo that welcome, Jillz.
In simple terms, anything traveling through a medium requires a certain amount of energy to do so, as it has to push stuff out of the way in order to progress. The denser that medium is, the more energy is needed because it's harder to push. (Think of swinging your arm around under water as opposed to in air.) Something in ballistic flight (ie: unpowered after launch) will run out of momentum sooner and thus start back down closer to the launch site.
Even without gravity, the trajectory will change in that it will take longer to get to where it's headed or even stop completely before getting there, even though the actual flight path doesn't change.

 

[Danger]

cute pict!

Thanks; it's clipped from my wedding photo.

 

[jillz]

With all of the calculations/data below, is it still possible to calculate the dimensions for the motorcycle falling/landing from the jump? Or is more data needed? What type of calculation would be used to find the dimensions for the total height the stuntrider/cycle will fall before hitting the ramp.Thanks for any help! Jillz


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Stunt man Joe wants to cross the Grand Canyon on his motorcycle. His crew has set up a ramp on one side of the canyon. Several factors will determine his safe landing on the other side. Assume that he decides to perform the stunt on a quiet day so that the drag from the wind will not have any effect on his motion.

Assume that Joe has to cross a distance of 40 m, leaving some clearance on both sides. The clearance on either side will be 6 m. Total horizontal distance that he needs to cover is 52 m to safely land on the other side.

The crew decides to set up the ramp at an inclination of 45o.
i.e. the slope of the ramp, θ = 45o

His motion from the time he leaves the ramp to the point where he is at the same level on the other side of the canyon. During the time he is airborne, his acceleration is the acceleration of gravity.

Using s = ut + ½ a t2
s = arc length

Horizontally: 52 = u cos θ t

t = 52 / u cos θ ------------------------------- (1)

Net vertical distance he traveled = 0

Vertically: 0 = u sin θ t - ½ g t2

u sin θ = ½ g t

Substituting from (1)

u sin θ = ½ g (52 / u cos θ)
u2 = 52g/ 2 cos θ sin θ = 52g/ 2 cos 45 sin 45 = 509.6

u (speed) = 22.6 m/s

He should reach a speed of 22.6 m/s (49.7 mi/h) at the end of the ramp in order to safely cross the Grand Canyon.

 

[Shooting Star]

Using s = ut + ½ a t2
s = arc length

The 's' above is not the arc length; it is the vertical distance traveled in time 't'. You are writing arc length, but applying the same formula for vertical displacement!

The rest seems to be all right, but I haven't checked the arithmetic.

 

[jillz]

So I would use the calculation: s = ut + ½ a t2
to find the distance the rider/bike would fall from the highest point down to the ramp??

 

[ZapperZ]

If things don't appear to make sense in here, that's because two different threads have been merged into one.

jillz: please do not make multiple threads of the same thing. Confine this question to ONLY this thread.

Zz.