a diver on a board 7.5m above the water walks off the end with a horizontal velocity of 2.3 m/s when they hit the surface of the water how far away from the edge of the board will they land?

2.846m

Explanation:

The diver is performing projectile motion.

To find x (final), we are going to use the equation x (final) = v (initial) * t + x (initial)

x (initial) = 0

x (final) = ?

v (initial) = 2.3 m/s

we don't know t

To find t we will use y (final) = 1/2 * (-9.8) * t^2 + v (initial in the y dir.) * t + y (initial)

- 9.8 in the acceleration in the y dir.

y (final) = 0

y (initial) = 7.5

v (initial in the y dir.) = 0

If we solve for t we get: t = 1.237s

Now we have all the components to solve for x (final) in x (final) = v (initial) * t + x (initial)

x (final) = 2.3*1.237 + 0

x (final) = 2.846m