A diver leaps from the edge of a diving platform into a pool. At time t seconds after she leaps, the horizontal distance from the front edge of the platform to the diver's shoulders is given by x(t), and the vertical distance from the water surface to her shoulders is given by y(t), where x(t) and y(t) are measured in meters. Suppose that the diver's shoulders are 11.4 meters above the water when she makes her leap that dx/dt=0.8 and dy/dt=3.6-9.8t, for 0<or= t <or= A, where A is the time that the diver's shoulders ender the water.
A) Find the max vert. distance from the water surface to the diver's shoulders.
B) Find A, the time that the diver's shoulders enter the water
C) Find the total distance traveled by the diver's shoulders from the time she leaps from the platform until the time her shoulders enter the water.
D) Find the angle θ, 0 < θ < л(Pi)/2, between the path of the diver and the water at the instant the diver's shoulders enter the water.
D = Sqaure root of ((dx/dt)² + (dy/dt)²) dt
The Attempt at a Solution
3.6 - 9.8t = 0
3.6t - 4.9t² + 11.4 = 0
t = 1.9362
square root of ((.8)² + (3.6 - 9.8t)²) dt
and i'm not sure how to finish this...
I don't even know where to start to try and figure this question out.