Projectile Motion of a Pebble on the Sloping Faces of Pyramid Cheops

[joemama69]

Homework Statement

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

Homework Equations

The Attempt at a Solution

y = v_o + v_yt + .5at
y = vtsin(90 - theta) + 4.9t


what else can i plug in. I am lost

 

[rl.bhat]

1.

The Attempt at a Solution

y = v_o + v_yt + .5at
y = vtsin(90 - theta) + 4.9t


what else can i plug in. I am lost

The above equations are wrong. Check them.

 

[berkeman]

Homework Statement

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

Homework Equations

The Attempt at a Solution

y = v_o + v_yt + .5at
y = vtsin(90 - theta) + 4.9t


what else can i plug in. I am lost

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...

 

[joemama69]

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

x = v_ocos(90-\theta)t

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

 

[joemama69]

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

 

[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