For those who can do a bit of algebra, the required equation of motion here is v^2 = u^2 + 2as assuming the short times involved don't give air resistance much time to affect the result.
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Here v is the final speed, u is the starting speed, a is acceleration which here is the gravitational acceleration at sea level, often given as 9.81 m/s/s, and s is the distance travelled. At the top of the jump, the starting vertical speed is zero, so u can be ignored.
Assuming the diver doesn't jump more than 1.5 metres up in the air, then s is between 3 and 4.5 metres for the lower board.
So v^2 is between 2x9.81x3 and 2x9.81x4.5, or 58.86 and 88.29 m^2/s/s.
Taking the square root gives v between 7.67 and 9.4 m/s (about 17 to 21 mph)
Assuming the diver doesn't jump more than 1.5 metres up in the air, then s is between 10 and 11.5 metres for the higher board (about the height of a three storey building or two giraffes).
So v^2 is between 2x9.81x10 and 2x9.81x11.5, or 196.2 and 225.63 m^2/s/s.
Taking the square root gives v between 14.0 and 15.0 m/s (about 31 to 33.6 mph)
Other equations can tell you that the highboard divers are falling for 1.4 to 1.5 seconds, while lowboard divers only have 0.8 to 0.95 seconds in which to do their tricks.
https://www.ncl.ac.uk/webtemplate/ask-assets/external/maths-resources/mechanics/kinematics/equations-of-motion.html#:~:text=at2.-,v%20%3D%20u%20%2B%20a%20t%20%2C%20s%20%3D%20(%20u%20%2B,represents%20the%20direction%20of%20motion.
