A stone is thrown horizontally with an intial speed of 8.0 m/s from the edge of the cliff. A stopwatch measures the stone's trajectory time from the top of the cliff to the bottom to be 3.4 s . What is the height of the cliff?
Thanks
2007-09-13 20:01:15 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
s=(1/2)at² and a=9.8 (on Earth) so
s=(1/2)*9.8*(3.4)² = 56.64 meters.
The horizontal component of its motion has nothing to do with how long it takes to fall. But it -does- contribute to where it lands. In this case it will land 3.4*8=27.2 meters from the base of the clif.
HTH
Doug
2007-09-13 20:06:55 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋
The initial speed doesn't matter because it is in the horizontal direction, and you are only asked about the vertical drop (height of cliff).
The vertical drop is given by
drop = 1/2 a t^2
where a = g = 9.8 m/s^2 and they give you t.
Plugnchug.
2007-09-14 03:08:01 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Here you should use the following eqn of kinematics
h = -ut + (1/2) g t^2 where h is the height of the clip
so, h = - 8 * 3.4 + 0.5 * 9.81 * 3.4^2
= -27.2 + 56.70
= 29.50 m
i hope that helps.
2007-09-14 04:09:32 · answer #3 · answered by Ehsan R 3 · 0⤊ 1⤋