# Projectile motion of a pebble

1. Sep 3, 2009

### joemama69

1. The problem statement, all variables and given/known data

a tourist is climbing high up the pyramid of cheops, which has sloping faces that make and angle of theta with the ground. The tourist throws a pebble with initial speed v is a direction perpendicular to one of the faces. the the height at which the peble hits the pyramid below the tourist

2. Relevant equations

3. The attempt at a solution

y = vo + vyt + .5at
y = vtsin(90 - theta) + 4.9t

what else can i plug in. im lost

2. Sep 3, 2009

### rl.bhat

The above equations are wrong. Check them.

3. Sep 3, 2009

### Staff: Mentor

Draw a sketch of the problem. Think both in the x and y directions as you write your equations. The pebble will have a constant velocity in the x direction of ____? Given that info, you can calculate how far down the face the pebble will hit as a function of time.

Then think in the y direction, and write an equation that describes the height y above the launch point the pebble will be as a function of time. Equate the two y's, and that's where and when the pebble will hit the face...

4. Sep 9, 2009

### joemama69

vx = vcos(90-$$\theta$$)
vy = vsin(90-$$\theta$$)

x = vocos(90-$$\theta$$)t

is this usefull tan$$\theta$$ = y/(vcos(90-$$\theta$$)t)

5. Sep 10, 2009

### joemama69

it seems i have 4 variables, x,v,t$$\theta$$ can i get a clue on how to eliminate them

6. Sep 10, 2009

### joemama69

ok made some progress but still a little confused

x = vtcos(90-$$\theta$$) therefore t = x/(vcos(90-$$\theta$$))

y = vsin(90-$$\theta$$))(x/(vcos(90-$$\theta$$))) - 4.8(x/(vcos(90-$$\theta$$)))

y = xtan(90-$$\theta$$)) - 4.8x/(vcos(90-$$\theta$$))) projectile

now i found the y based on the angle

tan$$\theta$$ = -y/x, therefore y = -xtan$$\theta$$ so i set both y's equal

-xtan$$\theta$$ = xtan(90-$$\theta$$)) - 4.8x/(vsin$$\theta$$)

is this correct so far

Last edited: Sep 10, 2009
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