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Bouncing bomb,

  • Thread starter qubert
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  • #1
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Question is:
plane flies at a speed of 100m/s at a height of 18m above water
plot the trajectory from release to first impact point, in intervals of 0.2m

managed this and have plotted the trajectory, with relevant velocities and resultant but the next part asks:

"At each bounce the vertical and horizontal velocity is reduced by a factor of 0.696, if the resultant speed has to be between 5 and 6m/s determine the number of bounces and horizontal distance the bomb requires without bouncing over."

how would one rearrange the equations finding the first trajectory without knowing the height the second bounce starts from?

i thought maybe by using a quadratic and pythagoras:

(100x0.696B)^2+(18.78x0.696B)^2=5.5^2 where b is the number of bounces, 100 and 18.78 are the horizontal and vertical velocities on the first drop

is this anywhere near correct?

thanks
 
Last edited:

Answers and Replies

  • #2
mgb_phys
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Firstly just consider the vertical direction.
Assume all the potential energy turns into kinetic at the ground.
Calculate the potential energy at the start and the kinetic energy at impact to get the vertical velocity at impact
Then use the new velocity to get the kinetic energy that it starts back up with.
Use this to get the potential energy and so the height of the next bounce.

Once you have the height and so the time for each bounce you can also get the horizontal velocity for each bounce in the same way.

I woudl probably do this a line at a time eg on a spreadsheet rather than trying to write an analytical eqaution for the end result.
 
  • #3
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is it possible to work out the potential energy without the mass of the object? would there be another way to work out the height of the next bounce?
 
  • #4
mgb_phys
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The mass is in the potential energy and the kinetic energy so it cancels out.
If the only change is in the velocity then you can only work it out with kinetic energy, of course you can write the an equation for the height of the next bounce but that is just using ke/pe anyway
 
  • #5
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so pe=176.4j before being dropped and ke=176.4j at impact, how would it be rearranged to find out the new velocity? (ive never used these formulae before so sorry if im slow) is the co efficient of restitution used here at all? thanks
 
Last edited:
  • #6
mgb_phys
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ke = 1/2 mv^2 so V = sqrt( 2 ke / m )
Then the V immediately after the bounce is 0.696V or you can use the equation to calcualte what the change in KE would be directly without workign out the speed.
 
  • #7
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sorry im still not following, would i be able to send you the excel spreadsheet i have so far so you can take a look at what ive done? thanks
 

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