Stone thrown question

  • Thread starter miamiheat5
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  • #1
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A stone is thrown vertically downward from a 2.00x10^2m high cliff at an initial velocity of 5.00m/s. (a) How long does it take for the stone to reach the base of the cliff? (b) What is the stones final velocity?


a) d = vi(t)+1/2a(t)^2
200=5(t)+1/2(9.81)t^2
0= -200 + (5)t + (4.910^2
4.91(t)^2 + (5)t -200
t= -5+- sqaureroot sign (5)^2-4(4.91)(-200)
-5+- squareroot sign 25+3953 / 9.81
t=6.0s

so i used the quadratic fromula

b) vf^2=vi^2+2ad
vf^2=5^2+2(9.81)(200)
vf^2=25+19.62(200)
vf^2 = 3949
vf= 63m/s
 

Answers and Replies

  • #2
960
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If you're just looking for a homework check station, you at least need to explain some uncertainty in reasoning. I can't speak for everyone here, esp the PF mentors.

For this problem, 63m/s as a dbl check should equate to the velocity to that gained by freefall in 200m less the distance from the initial velocity, then added to the initial velocity. If we take 6*5 and subtract from 200= 170 for the distance of the grav assist. We know that v=sqrt(2*a*y)
=sqrt (2*9.81*170), I get 57.75. Add to that 5, its close.
 
  • #3
Yeah, its correct.
 

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