Physics HW Help

Please help me with my physics! I am having a very hard time. It is just introductory physics but I am having difficulty understanding what to do.

Can you help me?

You are standing at the top of the cliff that has a stairstep configuration. There is a vertical drop of 6m at your feet, then a horizontal shelf of 10 m, then another drop of 4m to the bottom of the canyon, which has a horizontal floor. You kick a .93 kg rock, giving it an initial horizontal velocity that barely clears the shelf below. The acceleration of gravity is 9.8 m/s^2. Consider air friction to be negligible. What initial horizontal velocity LaTeX graphic is being generated. Reload this page in a moment. will be required to barely clear the edge of the self below you.

You are standing at the top of the cliff that has a stairstep configuration. There is a vertical drop of 6m at your feet, then a horizontal shelf of 10 m, then another drop of 4m to the bottom of the canyon, which has a horizontal floor. You kick a .93 kg rock, giving it an initial horizontal velocity that barely clears the shelf below. The acceleration of gravity is 9.8 m/s^2. Consider air friction to be negligible.
What initial horizontal velocity LaTeX graphic is being generated. Reload this page in a moment. will be required to barely clear the edge of the self below you.

The course I took was really a math problems course in disguise.  They don't seem to care if you know any theory.

>>4
protip: physics is just a bunch of algebra word problems.  sometimes calculus.

You need to consider horizontal and vertical seperately. First use the suvat equations to figure out out long it would take to drop the first 6m (treating it as a straight vertical drop), and then use the time you find in a second calculation to figure out u, the initial velocity of the horizontal kick.

>>2
The correct answer is 'Ow, bloody hell! My foot!'
Otherwise it's just reading comprehension with a tiny sprinkle of newtonian physics; how fast must an object go horizontally (at constant speed) to go 10m in the same time it vertically drops 6m with initial velocity 0, and 9.8m/s² acceleration downwards, if I read it right.
The given information is mostly superfluous.

Correct me if I'm wrong but I think this is the correct formula to how fast it drops, and going out would be the same rate divided by 2. Once again, I believe this is it:

9.8^2 = 96.04

9.8^2 * .93 = 89.32 / 10 = 8.932m/s for rate of fall

8.932 / 2 = 4.466m/s for rate of distance.

>>8
You can't get 4 digits of precision from 2 digits of inputs.

9.8² x .93 / 10 / 2 = 4.5

>>9
Clarification:  You can't get more digits of precision when using multiplicative methods.  If you added numbers together, then extra digits of precision can be achieved eventually.  For example:

9.8 x 9.8 = 96. = 2 digits of precision
9.8 + 9.8 = 19.6 = 3 digits of precision

Note also that the LEAST NUMBER of digits of precision governs the reported result.  If you multiply a 2-digit number by a 4-digit number, the result should be reported to only 2 digits of precision (unless the 2-digit number is exact).

Google "significant digits" for more information.

Who cares about significant digits anymore except Chemistry fags.

I have no idea how to solve this without a picture as a reference--and additionally, I believe you didn't even bother to include the entire problem--but remember that there is no horizontal component of acceleration in a problem like this:
Accel(Hor) = 0

therefore, Velocity(initialHorizontal) = Velocity(finalHorizontal)

Meaning, the graph of horizontal velocity would be a straight line at whatever value you started from. I don't know even know if that's what you're asking though.

>>10

In the simplest sense, yes you can gain digit count through addition, but with slightly more advanced analysis you would notice that the error would accumulate for repeated addition.

i.e.  2.4+5.3+6.2+5.1+5.0=24.0, but the error of order 10^-2 has probably accumulated to the point that we can really only say 24 instead of 24.0

This is why I'd prefer the use of standard deviations included with calculations rather than reliance on sigfigs e.g. 12.53±.61 instead of 12.5

>>6
Wow, now that I finally understand the question, I realize you posted the correct method a long time ago.