Projectile Motion - Time in Air

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To determine the time an arrow is in the air when launched at 60 m/s at a 30-degree angle and hitting the ground 1.8 m below its launch height, various attempts were made using projectile motion equations. The first attempt calculated time as 6.1 seconds, while the second attempt yielded 0.61 seconds using the height difference. The third attempt, which involved calculating the initial vertical velocity, resulted in a time of 0.86 seconds. The discussion highlighted the importance of considering the vertical displacement when calculating time in the air. Ultimately, the correct approach must account for the downward displacement of 1.8 m.
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Homework Statement


What angles should an arrow be launched if it leaves the bow @ speed of 60 m/s and is trying to hit a target that is 150 m away? Assume target is same height as bow.
Suppose the arrow is fired at an angle of 30 degrees & misses the target & hits the ground 1.8 m below from where it started. How much time was it in the air?

Homework Equations


1. t = 2vsinθ/g
2. t = √2y/g
3. Δy = (vsinθ)^2/2g

The Attempt at a Solution


I have attempted this the bold section of this problem a few different ways, but I am not convinced that any of them are correct. Could someone please tell me if any of them are correct, and if not, explain to me how to do it.

Attempt 1: t = 2(60)sin30/9.8
t = 6.1s

Attempt 2: t = √2(1.8)/9.8
t = .61s

Attempt 3: 1.8 = (vsin30)^2/2(9.8)
v = 8.4 m/s
t = 2(8.4)sin30/9.8
t = .86s
 
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Have you remembered to take into account the fact that the arrow lands 1.8m below it's original launch height?
 
_N3WTON_ said:
Have you remembered to take into account the fact that the arrow lands 1.8m below it's original launch height?
Wouldn't that be Δy, which I tried using in attempts 2 and 3?
 
Now that you have edited your solution I believe that you are correct..
 
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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