Summersnow8
  • Summersnow8
John stands on the edge of a deck that is 40.0 m above the ground and throws a rock straight up that reaches a height of 15.0 m above the deck. What is the initial speed of the rock in m/s? which i solved correctly and it is 17.1 Assuming the rock in the previous question misses the deck on the way down, how fast (in m/s) will the rock be moving when it hits the ground? @ybarrap @lightgrav
Physics
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SOLVED
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katieb
  • katieb
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Summersnow8
  • Summersnow8
i used the equation \[2as = V _{f}^{2} - V{i}^{2}\]
anonymous
  • anonymous
so, use that correct answer in the same statement (equation) as before, but solve it for v_f ... remember that (now) a is in the same direction as s (both downward) so 2as is positive.
anonymous
  • anonymous
(or, you could start with v_i = 0, but watch the rock fall 55m

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Summersnow8
  • Summersnow8
so, \[2 (-9.8)(-40) = Vf ^{2} - (17.1)^{2}\]
anonymous
  • anonymous
looks right so far
Summersnow8
  • Summersnow8
so vf would be 32.8?
anonymous
  • anonymous
yes, 55m is almost 4x the upward 15m (so v should be almost double)
ybarrap
  • ybarrap
You got it! Another way to look at this - $$ V_f^2 = V_i^2 + 2gs $$ Where \(V_i=0\), the speed of the rock at the highest point. \(s\) is is the distance the rock is above the deck plus the height the deck is above the ground: 15+40=55. Now you have everything to solve. You've got the right approach either way!
Summersnow8
  • Summersnow8
so is 32.8 the answer? i was told to use the equation i listed
anonymous
  • anonymous
His eq'n is the same as yours, just re-arranged for v_f
ybarrap
  • ybarrap
Yep! $$ V_f^2 = V_i^2 + 2gs=0-2\times9.81\times55\\ \implies V_f=32.8~m/s $$
ybarrap
  • ybarrap
*Negative, because rock is going down: $$ V_f^2 = V_i^2 + 2gs=0-2\times9.81\times55\\ \implies V_f=\color{red}{-}32.8~m/s $$
Summersnow8
  • Summersnow8
..... negative?
ybarrap
  • ybarrap
Just indicates direction, downward. Up is positive
anonymous
  • anonymous
the same direction as the acceleration.
Summersnow8
  • Summersnow8
it says it's wrong
anonymous
  • anonymous
ooh, they asked for _speed_ (how fast) not velocity :-(
ybarrap
  • ybarrap
speed is magnitude, velocity includes directional information. So speed would just be 32.8 m/s

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