What is the velocity of a body thrown into a pit?

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A body is thrown upward from a height of 1.5 m above the ground into a pit that is 3.5 m deep, resulting in a total height of 5 m. The initial velocity of the body is 2.3 m/s. Using the equation V^2 = Vo^2 - 2gy, the final velocity at the bottom of the pit is calculated to be approximately 10.16 m/s downward. Participants in the discussion confirm the calculations and clarify the importance of considering the initial height in the equation. The problem highlights the complexities involved in projectile motion and energy conservation principles.
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Homework Statement


A body is thrown vertically upward from the distance h = 1,5 m above the ground at the
edge of a pit having a depth of s = 3,5 m. The initial velocity of the body is 2,3 m/s.

Homework Equations


Determine the velocity at which the body will reach the bottom of the pit.

The Attempt at a Solution


(h=5m)

V2-V02=2gh
V=10.168 m/s
 
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What a violent problem...

I'll give you a hint, the initial height is 1.5m, but the total height is 3.5m.
 
flyingpig said:
What a violent problem...

I'll give you a hint, the initial height is 1.5m, but the total height is 3.5m.

really ?
 
chawki said:

The Attempt at a Solution


(h=5m)

V2-V02=2gh
V=10.168 m/s
Looks good.
 
Hello Doc Al, Thank you for your help on the last time regarding equations..i try not to mess up my mind with them.

regarding this actual problem, I'm starting to think that flying pig is right :/
 
chawki said:
regarding this actual problem, I'm starting to think that flying pig is right :/
Why is that?
 
Take Doc Al's advice. I just glanced over the problem and your equation. I just saw that you missed the initial height in your equation, so I jumped into a conclusion lol
 
flyingpig said:
Take Doc Al's advice. I just glanced over the problem and your equation. I just saw that you missed the initial height in your equation, so I jumped into a conclusion lol

ahaa..lol don't confuse me :smile: I'm easy to confuse :shy:
i'm glad i was right at the first try :smile:
 
This is how I'm imagining it...
 

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  • #10
Am i right ?
 
  • #11
chawki said:
Am i right ?
Yes.
 
  • #12
Doc Al said:
Yes.

Thank you :smile:
 
  • #13
i think you're right boy .. here's my solution

v^2 = Vo^2 - 2gy
Vo=2.3m/s
y=1.5+3.5 = 5m

V=2.3^2 -2(9.8)(-5)
V=10.16m/s downward ..

:approve:
 
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