Calculating Vertical Distance of a Thrown Pebble from a Pyramid

[joemama69]

Homework Statement

A pebble is thrown off the side of a pyramide perpindicular to its slope. Find the vertical distance h under the assumption the drag forces are negligible.

Homework Equations

The Attempt at a Solution

v_x=vcos(90-Q) Q = theta
v_y=vsin(90-Q)

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x = v_xt; therefore t = \frac{x}{v_x} = \frac{x}{v_{x}cos(90-Q)}

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tanQ=\frac{y}{x}; therefore y = xtanQ

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y=y_o + v_yt - \frac{1}{2}gt²

y= \frac{xv_{x}sin(90-Q)}{v_{x}cos(90-Q)} - \frac{9.8}{2}(\frac{x}{v_{x}cos(90-Q)})²

y= xtan(90-Q) - \frac{4.8x^{2}}{v^{2}cos^{2}(90-Q)}


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Set both y's equal to each other

xtanQ = xtan(90-Q) - \frac{4.8x^{2}}{v^{2}cos^{2}(90-Q)}

tanQ = tan(90-Q) - \frac{4.8x}{v^{2}cos^{2}(90-Q)}

now I am going to solve it for x

x = \frac{v^{2}cos^{2}(90-Q)[tanQ-tan(90-Q)]}{-4.8}

and now i pluf it into y = xtanQ

y = \frac{v^{2}tanQcos^{2}(90-Q)[tanQ-tan(90-Q)]}{-4.8}

seems a little to messy, did i miss somethig

 

[rl.bhat]

It is right. But you can put it in a better form.
write tan(90 - Q) = cot Q
cos(90 - Q) = sin Q

 

[ehild]

Take care, the slope goes downwards so y = -xtanQ. And 9.8/2 =4.9.

ehild

 

[joemama69]

-tanQ = tan(90-Q)-\frac{4.9x}{v^{2}cos^{2}(90-Q)}

x = \frac{v^{2}cos^{2}(90-Q)}{4.9}[tan(90-Q) + tanQ]

x = \frac{v^{2}sin^{2}Q}{4.9}[cotQ + tanQ]

y = -\frac{v^{2}tanQsin^{2}Q}{4.9}[cotQ + tanQ]


so there is nothin left to do with this one

 

[rl.bhat]

In the last step put cotQ = 1/tanQ and simplify.

 

[joemama69]

y = \frac{-v^{2}sin^{2}Q[1 + tan^{2}Q]}{4.9}

 

[rl.bhat]

y = \frac{-v^{2}sin^{2}Q[1 + tan^{2}Q]}{4.9}

1 + tan^2θ = sec²θ