Posted by tommy on Tuesday, September 7, 2010 at 1:45pm.
REPOST PLEASE HELP a rock is thrown horizontally with a speed of 17 m/s from a vertical cliff of height 34 m.
A I solved for t=2.63
BHow far will it land from the base of the cliff?
im using x=xo+vox*t
i get 6.46 and it is wrong. What am I doing wrong?
CWhat is the velocity (magnitude and direction counterclockwise from the +xaxis, which is the initial horizontal direction in which the rock was thrown) of the rock just before it hits the ground?
Magnitude m/s
Direction °
what would i use for a formula?

physics  drwls, Tuesday, September 7, 2010 at 1:55pm
Your time is correct, but they want the horizontal distance from the base opf the cliff. How can that be negative?
X = 17 m/s*2.63 = 44.8 m
Your formula for x is crrect, but xo = 0 and I don't see how you came up with 6.46.
For the speed when it hits the ground, you can use conservation of energy.
[Vfinal^2  Vinitial^2]/2 = g H
For the direction,
Vinitial/Vfinal = cos^1 theta
Vinitial = Vx (during flight)= 17 m/s
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