Finding time and speed at which skydiver hits the ground

[andrey21]

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!

 

[rock.freak667]

How about using v=u+at?

 

[andrey21]

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?

 

[andrey21]

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

 

[andrey21]

Anymore ideas please, help desperatley needed!

 

[Dick]

Looks ok to me if the initial height is 4000m. 282m/s IS fast, but you've completely neglected air friction. That would slow the skydiver down a lot, even if the chute didn't open.

 

[andrey21]

Ah yes thanks for the post :)