a rock is thrown at a cliff at a velocity of 60 feet per second at an angle of 45 degrees for a distance of 80 feet.
What height does the rock hit the cliff?
Use 32 ft/s^2 as g.

I know this can be done with physics but i am looking for a way to solve this with calculus

Well you could always "forget" the kinematic equations, then use calculus to derive them:

do you mean using a lagrangian?

or this? http://en.wikipedia.org/wiki/Principle_of_least_action

I'm pretty sure you can use the Tangent Trig Identity to solve the problem.

Make a right triangle fill in given information: distance traveled = 80 ft. Angle = 45 deg., in a right triangle the angle opposite the hypotonuse is 90 deg.

Label:
Angle A - 45 deg
Angle B - 45 deg
Angle C - 90 deg

Label:
Leg a (opposite angle A) - ?
Leg b (opposite angle B ) - 80 ft
Leg c (opposite angle C) - ?

leg a - denotes height of cliff
leg b - denotes distance traveled from starting point to ending point
Leg c - hypotenuse

to find side a, we can use a Tangent trig Identity, Tan(a) = opposite leg / adjacent leg, so we get:

Tan(45) = a / 80 ft
a = 80 x Tan(45) deg
Tan(45) deg or Tan(pi/4) = root 2 / root 2 = 1

a = 80 x 1
a = 80 ft.

I dunno though, I forget about the velocity equations and how it might change the equation.

This post was edited by Jeffry on Oct 29 2013 03:10pm

yeah that would be the answer if there were no accelerations

Ok I haven't done this in a very long time so I'm not sure if its a 100% correct -

you would use the equation 60(t)sin(45deg)-4.9(t)^2
t = time
Then figure out the time it takes to travel 80 ft horizontally, i came up with the answer of 1.581 seconds.
Plugged it into the equation
60(1.581)sin(45deg)-4.0(1.581)^2 = 54.8261492634 ft high.