How to find the final height of a moving object

  • Thread starter Thread starter lordloss
  • Start date Start date
  • Tags Tags
    Final Height
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 6K views
lordloss
Messages
7
Reaction score
1

Homework Statement



A startled armadillo leaps upward, rising 0.523 m in the first 0.206 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.523 m? (c) How much higher does it go?


Homework Equations



The five Kinematic equations

The Attempt at a Solution



I worked out A and B and got the following answers:

(a) 3.54 m/s

(b) 1.5 m/s

I just can't figure out the last part, C, on how to find how much higher the armadillo goes. Can anyone help me out with this?
 
Physics news on Phys.org
I would use [tex]2a(x-x_0)=v^2-v_0^2[/tex]. Do you see how you can use this to solve (c)?
 
I'm using WileyPlus and its not taking any of the answers I have come up with.

I inserted all the variables into this modified equation to find the final x variable:

X=(V^2-V0^2+2aX0)/2a

Did I do the algebra wrong?

I keep getting 1.047m.
 
That's wrong, and it looks like you are plugging in the wrong velocities here.

For part (c), the armidillo's "initial" velocity is the 1.5 m/s it has at the 0.523 m height. It's "final" velocity is the velocity it has when it has reached it's maximum height (and that velocity is ____?)