# Finding time and speed at which skydiver hits the ground!

If initial vetical velocity is zero, at what time and how fast does a skydiver hit the ground? Assume that g=10m.s^-2

Given that:
y' = -gt + A y' is the vertical velocity
y = -(1/2).g.t^2 +At +B y is the distance above the ground

In order to solve how long takes skydiver to hit ground I did the following:

0 = -(1/2).g.t^2 +(0)t + 4000
4000 = -(1/2).g.t^2
8000 = g.t^2
800 = t^2
t = 28.284 seconds!!!

Now from here I am unsure how to find the speed at which hits the ground. Help needed sorry for bad formatting!

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rock.freak667
Homework Helper

Okay so if initial velocity is zero I get:

v = 0 +10.28.284
v = 282.84 m.s^-2
Is this correct?

It seems bit excessive as 282.84m.s^-2 is 632.319 MPH. DOes that seem correct to you?

I must also say the skydivers parachute fails to open so a higher speed on descent would be expected.

Anymore ideas please, help desperatley needed!

Dick