What Are the Kinematics of Cliff Divers in Acapulco, Mexico?

[Peter G.]

The cliff divers of Acapulco, Mexico, take off horizontally from a rocky cliff face 26.5 m above the surface of the water. In the course of their flight, of which all but the last 1.5 m of their horizontal motion is above rock, they travel 8.0 m forward:

a) For how long are they in the air (ignoring air resistance)
b) For how long are they over the water
c)What is their vertical velocity on entry
d)At what angle is their path to the vertical at entry?

I attempted all of them but I am not sure about some of:

a)h = 1/2gt^2
26.5 = 5t^2
t = 2.30 s

b) I did Direct Proportion: If it takes 2.3 seconds for 8 m: Would be 0.432 s over water

c) v^2 = u^2 + 2as
V^2 = 530
v= 23.02 m/s

d) For the last one I used 23.02 m/s for the vertical velocity and for the horizontal (8 m / 2.3 s = 3.47) and calculated the angle

tan (3.47 / 23.02) = 8.58 degrees

(I did not use the rounded values for the calculations, I just didn't write them down here)

Can you guys help me with these questions please?

Thanks,
Peter

 

[collinsmark]

Well, besides the issue of significant figures, which you've already addressed, everything looks correct to me.

(And of course, I'm assuming that you were instructed to use 10 m/s² for g, as opposed to 9.8 or 9.81 m/s². Using 10 m/s² for g when you're supposed to use one of the more precise values, might have a significant impact on the final answer.)

 

[Peter G.]

Yup, gravity was 10.

Thanks!