Conceptual problem involving ME and intelligence

i lack the latter, so here is my question, if three objects are shot from a cliff all at different angles.....what will their KE be right before they hit the ground relative to each other (neglecting air resistence) now i know the answer to this is they are all equal, having gotten it wrong on a test, but lets say the object shot at the highest angle, doesnt that one have the furest to fall meaning more PE due to gravity,,, how can they possibly all have the same KE,, ie speed. The lame answer i got from my professor was that they all started at the same height, but the object shot at the highest degree angle was at a different height, so it has more PE....confused and frustrated
 

Claude Bile

Science Advisor
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The KE of all the objects are indeed equal, assuming that all the objects have the same initial speed.

Final KE = Initial KE + Initial PE.

This is just conservation of energy. If all objects fall the same distance and start with the same amount of kinetic energy, then conservation of energy demands that the objects' final KE be equal.

I hope that makes sense.

Claude.
 
164
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The object shot at the highest angle has higher potential energy at the top of its flight path than the other two objects. However, the objects shot at lower angles are moving faster horizontally, which makes up for it. That's why they all have the same KE.
 
164
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Just to be clear I should say they all have the same KE when they reach the ground and PE is equal zero.
 
Well, if i shoot them all at different angles, the steepest angle gets the most height, but also has the least horizontal component of velocity. When it hits the ground lets assume that the state of potential energy at this height is zero. Thus the kinetic energy is simply 1/2m*v DOT v. v DOT v is simply v^2 but is a scalar. v = sqrt(vx^2 + vy^2) so then vf^2 = vxf^2 + vyf^2 if we break it up into components and assume 2-D motion. vxf = v0cos(theta), vyf = v0sin(theta) - gt. Lets assume the ball lands on the height it was launched.
vf^2 = vo^2cos(theta)^2 + vo^2sin(theta)^2 -2*voy*g*t + (g*t)^2

but for parabolic motion w/ the same landing height we have
0 = voyt - gt^2/2 or that voy = gt/2 or (2*voy) = g*t
this means

vf^2 = vo^2cos(theta)^2 + vo^2sin(theta)^2 -2*voy*g*t + (g*t)^2
vf^2 = vo^2cos(theta)^2 + vo^2sin(theta)^2 -g*t*g*t + (g*t)^2
vf^2 = vo^2cos(theta)^2 + vo^2sin(theta)^2 -(g*t)^2 + (g*t)^2
vf^2 = vo^2cos(theta)^2 + vo^2sin(theta)^2
vf^2 = vo^2(cos(theta)^2 + sin(theta)^2)
vf^2 = vo^2

KEf = KE0 assuming mass is constant
 

HallsofIvy

Science Advisor
Homework Helper
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But the potential energy at the top point is irrelevant to the problem. They all have the same kinetic and potential energy initially and so will have the same kinetic energy just as they hit the ground.
 
thanks to all for the quick and informative help
 

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