A cliff diver horizontally runs off a 10 m high cliff at 10 m/s. How far from the base of the cliff
does he land? (Round answer to nearest 10ths place)
Need help ASAP

We know it will take 1 second to reach the ground if we use (10m/s^2) as the acceleration due to gravity (use 9.81m/s^2 for regents physics) Therefore we can substitute into the equation d=Vi(t)+1/2(a)(t)^2

d=(10m/s)(1sec)+1/2(0m/s^2)(1sec)^2
so d=10m (adjust if using 9.81 m/s^2)

It helps to make an XY chart with the Vi, Vf, a, d, and t values for your vertical and horizontal components before attempting to solve anything.

Help this helps!

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way