Projectile motion problem

  • Thread starter david1111
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  • #1
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Homework Statement


j19219Q1n179.png



Homework Equations


R = 1/2 gt^2
R+x = vt




The Attempt at a Solution


For a), I tried to find t first, which is equal to √(2R/g). Then, I put t = √(2R/g) into the equation R+x = vt, and I found out that v = (R+x)/√(2R/g), is that correct?

For b), is the answer (R+x)?
 

Answers and Replies

  • #2
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hint: for the ball to reach the ground without hitting the rock, the condition is that the y-coordinate of ball should be greater than a certain height.

placing the origin at the center of sphere, ##R^2=x^2+y^2##
##(x,y)## are coordinates on the surface of hemisphere

use kinematic equation to determine the height of ball at each instant and replace ##t## from the equation.

taking ##x## as common coordinate, relate:

height ≥ ##y##

this is sufficient to solve the question. Try it.
P.S. My ##x## is different than the one asked!!!
 
  • #3
tiny-tim
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hi david1111! :smile:
For a), I tried to find t first, which is equal to √(2R/g). Then, I put t = √(2R/g) into the equation R+x = vt, and I found out that v = (R+x)/√(2R/g), is that correct?

For b), is the answer (R+x)?

yes, that's all correct, but you don't know what x is!

hint: what is the value of the centripetal acceleration? :wink:
 
  • #4
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Centripetal acceleration= v^2 / R? But, how to relate it and x?
 
  • #5
tiny-tim
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what is the value of it?

(you can find the value without knowing the speed)
 
  • #6
haruspex
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Hi david1111,

NihalSh and tiny-tim have offered two quite different methods. T-T's is easier but requires the assumption that if it is going to hit the rock again it will do so immediately. That can be justified by thinking about how the curvature changes in a parabola, but NihalSh's method avoids this. It would be quite instructive to do both.
 

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