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

B-How 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?

C-What is the velocity (magnitude and direction counterclockwise from the +x-axis, 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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