A diver leaves a 3-m board on a trajectory that takes her 2.8m above the board and then into the water 2.7m horizontally from the end of the board.At what speed and angle did she leave the board?

Answer 1

Formulas for projectile motion:
h,max = vo^2 sin^2 (x) / 2g
R = vo^2 sin (2x) / g
where
h,max = maximum height
R = range
vo = initial velocity
x = angle
g = acceleration due to gravity = 9.8 m/s^2
In the problem, we are asked to find vo and x, and we have two equation, so we can solve for them:
2.8 = vo^2 sin^2 (x) / 19.6 : equation (1)
2.7 = vo^2 sin (2x) / 9.8 : equation (2)

To solve this, what we can do is to divide equation (1) by equation (2):
2.8/2.7 = (sin^2 (x) / 19.6) / (sin (2x) / 9.8)
1.03704 = sin^2 (x) / 2*sin(2x)
Note that sin(2x) = 2 sin(x) cos(x), thus
1.03704 = sin^2 (x) / 2*2*sin(x) cos(x)
1.03704 = sin(x) / 4 * cos(x)
4.148148 = sin(x) / cos(x)
Note that sin(x)/cos(x) = tan(x), thus
4.148148 = tan(x)
x = 76.44 degrees

Now that you have a value for angle, substitute this back to either equation and solve for vo.
hope this helps~ `u`
(I apologize is someone already posted an answer/solution before me. My internet is really slow and I can't seem to post a comment right away.)