On a cliff edge!

  • #1
Lucy stands on the edge of a vertical cliff and throws a stone vertically upwards. The stone leaves her hand with speed 15m/s at the instant her hand is 80m above the surface of the sea. Air resistance is negligible and the acceleration of free fall is 10m/s/s.

First part is to calculate the maximum height reached which I've done and is 11m.

Second part is to find the time for the stone to reach the sea. I have found the correct answer of 5.8s eventually using the quadratic equation which is fine but quite tricky. This is where my question comes in.

The mark scheme also mentions as an alternative finding 4.3s as the time to fall 91m added to 1.5 for max height or finding 3 to return to hand and then 2.8 to fall 80 m. This is annoying me as I can't see where these answers come from? I assume the 1.5 comes from v/a = 15/10 and that it is doubled for 3 but how are they getting 4.3 and/or 2.8? I've tried every combination of motion equation and I'm sure I'm missing something obvious, please help!
 

Answers and Replies

  • #2
Borek
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At the top of the trajectory stone doesn't move, so it has to fall 91 m with acceleration g and no initial speed. That yields 4.3 s.

When the stone goes down, once it reaches the cliff level it has exactly the same speed, but opposite direction - so you have a free fall with v0 = 15 m/s (do you know why?) and 80 meters to go. That yields 2.8 s.

Up and down in 3 s is just a free fall with initial speed going up. Going up takes exactly the same time it takes to go down.
 
  • #3
adjacent
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When the stone goes down, once it reaches the cliff level it has exactly the same speed...
Not the cliff level,the hand level(80 m)
 
  • #4
Borek
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Not the cliff level,the hand level(80 m)

Good point, lousy wording on my side.
 

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