Finding How Far From the Ground an Object Hits Another

Giygas72

1. The problem statement, all variables and given/known data

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

2. Relevant 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

3. 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

gneill

Quote by Giygas72

1. The problem statement, all variables and given/known data

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

2. Relevant 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

3. The attempt at a solution

= 0 + 2(-9.8)(12) [itex]\color{red} \Leftarrow[/itex]
= (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?

SammyS

Quote by Giygas72

1. The problem statement, all variables and given/known data

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

2. Relevant 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

3. 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 v_f² = v_i² + 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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