What Is the Acceleration Needed to Stop an Arrow?

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To determine the acceleration needed to stop an arrow embedded 8 inches into the ground after being shot straight up at 220 ft/s, it's essential to focus on the interval from the moment it strikes the ground to when it stops. The acceleration is not constant throughout the entire motion due to changing forces once the arrow penetrates the ground. The final velocity when the arrow hits the ground is equal in magnitude but opposite in direction to its initial velocity. By treating the scenario as if the arrow were fired directly downward, the calculations simplify significantly. This approach allows for a straightforward application of kinematic equations to find the required acceleration.
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Homework Statement


An arrow is shot straight up in the air with an initial speed of 220 ft/s. If on striking the ground it embeds itself 8.00 in into the ground, find the magnitude of the acceleration (assumed constant) required to stop the arrow, in units of feet/second^2.

Homework Equations


4 basic kinematic equations.

The Attempt at a Solution


Does anyone see where I went wrong in my work?
 

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The problem is just that you're using constant acceleration over the entire distance from the top of its path down to when it's embedded in the ground, which you can't do since the acceleration changes once it's in the ground. There's some funky stuff in your equations for plugging in the initial and final velocities because of this. Try instead using the interval from when the arrow hits the ground to when it stops 8 in. later.
 
Becomes a really easy problem when you know Vf=-Vi for velocities.
 
or in other words.. if you throw something up in the air with velocity V it will hit the ground at the same velocity V. So you can forget that part of the problem and pretend the arrow was just fired at the ground.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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