• Support PF! Buy your school textbooks, materials and every day products Here!

Finding time and speed at which skydiver hits the ground!

  • Thread starter andrey21
  • Start date
  • #1
466
0
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!
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
How about using v=u+at?
 
  • #3
466
0
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?
 
  • #4
466
0
I must also say the skydivers parachute fails to open so a higher speed on descent would be expected.
 
  • #5
466
0
Anymore ideas please, help desperatley needed!
 
  • #6
Dick
Science Advisor
Homework Helper
26,258
618
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.
 
  • #7
466
0
Ah yes thanks for the post :)
 

Related Threads for: Finding time and speed at which skydiver hits the ground!

Replies
3
Views
3K
Replies
17
Views
751
  • Last Post
Replies
6
Views
4K
Replies
3
Views
2K
Replies
0
Views
920
  • Last Post
Replies
16
Views
5K
Replies
7
Views
10K
Replies
3
Views
3K
Top