A diver leaves a 3-m board on a trajectory that takes her 2.5 m above the board, and the into the water a horizontal distance of 2.8 m from the end of the board. At what speed and angle did she leave the board? Book answer is 7.2 m/s at 77 degrees to the horizontal.
y= y0 + vy0t - 1/2gt^2
y = x tan theta0 - g/2v0^2 cos^2 theta0*x^2
The Attempt at a Solution
If the origin is at water level, then at the top of the dive vy0 = 0,and 1/2/gt^2 = 5.5 m. Therefore, t = 1.06 s. I still have two unknowns, v0 and theta, but only one equation.