What is the height of the cliff?

[kchurchi]

Homework Statement

You want to know the height of a cliff so you throw a rock off the edge with initial speed v0 = 42 m/s upwards, inclined at an angle θ = 27° with respect to the horizontal. You have a friend that records the time that it takes for the rock to hit the bottom of the canyon below. It takes 7.6 s for the rock to hit the bottom of the canyon from when you throw it.

Homework Equations

v0x = v0*cos(theta)
v0y = v0*sin(theta)

ay = -g
ax = 0

vy(t) = v0*sin(theta) - g*t
vx(t) = v0*cos(theta)

x(t) = h + v0*sin(theta)*t - 1/2*g*t^2
y(t) = v0*cos(theta)*t

h = height of the cliff
v = velocity
v0 = initial velocity
a = acceleration

The Attempt at a Solution

I am just confused about how to find the height of the cliff. I can't use the initial time because the initial time is zero. I tried solving for the x-distance at the final time. but I don't know how that would help. Any suggestions please??

 

[jldibble]

You have your equations for the x and y position mixed up. You only need to worry about what's happening in the vertical (y) direction.

The position equation for the y direction is y = h + v0*sin(theta)*t - 1/2*g*t^2

In this equation, y = 0 because the rock will land on the ground level which is easiest to assume is zero.

t = 7.6s

Plug in everything else and it should work out to get you an answer of 138.6 meters (give or take).