Two divers jump from a 2.70meter platform. One jumps upward at 1.60meters/second and the 2nd steps off the platform as the first passes it on the way down. What are their speeds as they hit the water? Which hits the water first and by how much?
Vf^2= Vi^2 + 2ad (Vf being final velocity and Vi being initial velocity)
Y=Y0 + ViT - 0.5gt (Y and Ynaught being the distance in the y-axis direction, T being time, and g being gravity)
The Attempt at a Solution
So i used the equation Vf^2=Vi^2 + 2ad to find the speed at which the diver that walked off the platform hit the water:
Vf^2= 0 + 2(-9.8m/s^2)(0-2.7m) = sqroot[(-18.6m/s^2)(-2.70m)]= 7.09m/s
I'm somewhat sure that to find the speed of the other diver I have to use that 2nd equation that I provided and 1 more additional equation, but yeah, Im stuck. Can anyone help? Equations and explanations are much appreciated, thanks
Oh, and off the bat I'm going to say the diver that jumped from the platform will reach the water first as opposed to the diver that walked off. But I don't know how to show that, so yeah...