Calculating Maximum Range of a Catapulted Rock

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The discussion focuses on deriving the theoretical maximum range of a rock launched from a catapult, expressed as R=2Mh/m, where M is the mass of the counterweight, m is the mass of the rock, and h is the height from which the counterweight falls. Participants explore the relationship between potential energy (Ep) and kinetic energy (Ek) to establish this formula. The optimal launch angle for achieving maximum range is identified as 45 degrees. Additionally, there is inquiry into calculating the range when the launch angle is 45 degrees and the velocity is a variable v. The conversation emphasizes the physics principles governing projectile motion and energy conversion in catapult mechanics.
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Homework Statement


Assuming that the rock launched by a catapult is released from ground level, show that the theoretical maximum range is:
R=2Mh/m, where M = mass of counterweight
m = mass of rock
h = height counterweight falls


Homework Equations



Ep=mgh
Ek=(mv^2)/2

The Attempt at a Solution



I equated the two expressions and then split the velocity into its components so this splits the Ek up into horizontal and vertical but I can't progress from here
 
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At what angle should a projectile be launched to obtain a maximum range?
 
45degrees
 
What is the range when the angle is 45 degrees and the magnitude of velocity is some v?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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