Finding How Far From the Ground an Object Hits Another

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In the discussion, a physics problem involves calculating the height at which an arrow strikes a sandbag dropped from a 12-meter tower after 1.1 seconds and determining the arrow's initial velocity. The participant initially misapplies the kinematic equation vf^2 = vi^2 + 2g*t, which is incorrect for this scenario. Instead, the correct approach involves using the equation for distance fallen under gravity to find the height of the sandbag after 1.1 seconds. The conversation emphasizes the need to select appropriate kinematic equations that account for time and acceleration due to gravity. Ultimately, the focus is on correctly applying physics principles to solve the problem.
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Homework Statement



For archery practice, a knight's squire drops sandbags from a 12.0 metre tower. At exactly the same time the sandbag is dropped, the knight shoots an arrow up at the sandbag from the base of the tower. If the arrow strikes the sandbag at 1.1 seconds, calculate;

a) how far from the ground the arrow strikes the sandbag
b) the arrow's initial velocity


Homework Equations



I chose vf^2 = vi^2 + 2g*t equation, ended up with this...

Other equations:
vf = vi + g*t
d = 1/2(vi + vf)*t
d = vi*t + 1/2g * t^2


The Attempt at a Solution



= 0 + 2(-9.8)(12)
= (the square root of) 235.2
= 58m

Obviously incorrect. The biggest problem I'm having is finding the right equation to use. Where exactly do I go from here?
 
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Giygas72 said:

Homework Statement



For archery practice, a knight's squire drops sandbags from a 12.0 metre tower. At exactly the same time the sandbag is dropped, the knight shoots an arrow up at the sandbag from the base of the tower. If the arrow strikes the sandbag at 1.1 seconds, calculate;

a) how far from the ground the arrow strikes the sandbag
b) the arrow's initial velocity


Homework Equations



I chose vf^2 = vi^2 + 2g*t equation, ended up with this...

Other equations:
vf = vi + g*t
d = 1/2(vi + vf)*t
d = vi*t + 1/2g * t^2


The Attempt at a Solution



= 0 + 2(-9.8)(12) \color{red} \Leftarrow
= (the square root of) 235.2
= 58m

Obviously incorrect. The biggest problem I'm having is finding the right equation to use. Where exactly do I go from here?

What units are associated with the numbers on the line indicated with the arrow? What then should be the units of the result?

It would appear that you need to choose an equation that will give you the height of a sandbag at a given time t after it is dropped. Which of your Relevant Equations gives you distance with respect to time when acceleration is involved?
 
Giygas72 said:

Homework Statement



For archery practice, a knight's squire drops sandbags from a 12.0 metre tower. At exactly the same time the sandbag is dropped, the knight shoots an arrow up at the sandbag from the base of the tower. If the arrow strikes the sandbag at 1.1 seconds, calculate;

a) how far from the ground the arrow strikes the sandbag
b) the arrow's initial velocity

Homework Equations



I chose vf^2 = vi^2 + 2g*t equation, ended up with this...

Other equations:
vf = vi + g*t
d = 1/2(vi + vf)*t
d = vi*t + 1/2g * t^2

The Attempt at a Solution



= 0 + 2(-9.8)(12)
= (the square root of) 235.2
= 58m

Obviously incorrect. The biggest problem I'm having is finding the right equation to use. Where exactly do I go from here?

There is no such kinematic equation as
vf^2 = vi^2 + 2g*t .​

There is one which states that vf2 = vi2 + 2g*d , but it's not much good for solving part a .

How far does any object fall in 1.1 seconds, when dropped from rest ?
 
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