Why Did I Calculate the Cliff's Height Incorrectly?

  • Thread starter confusedaboutphysics
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In summary, using the given information of the diver's velocity and time, the cliff was calculated to be 44.1m high and the distance from the base to where the diver hit the water was 3.6m. The correct formulas were used, but there was a mix-up in the calculations.
  • #1
confusedaboutphysics
33
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need help please! what am i doing wrong??

A diver running 1.2 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3.0 s later. How high was the cliff and how far from its base did the diver hit the water?

so does v = 1.2 m/s and t=3 seconds
i used the formula Vxo=X/t so i did 1.2=x/3 so x = 3.6m for the first part of the question. for the second part of the question i used the formula y=-1/2gt^2 so i did y=(-1/2)(-9.8)(3^2) = 44.1m.
so i said that the cliff was 3.6m high and the base was 44.1m away. but i got it wrong. am i using the right formulas?
 
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  • #2
You got the right formulas but you switched em, he landed 3.6 m away and the cliff was 44.1m high!
 
  • #3
ohh! thank you so much! it didn't sound right that a cliff was 3.6m high and he hit the water at 44.1m! lol
 

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