Is 570 Meters a Correct Estimate for the Balloon's Height?

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SUMMARY

The discussion centers around the calculation of a hot air balloon's height, estimated at 570 meters, based on the horizontal distance a dropped bag lands from a boat. The user applied the horizontal motion formula, x - x0 = (v0 cos(theta))t, and the vertical motion formula, y - y0 = (v0 sin(theta)) - 1/2 gt^2, to derive the height. Concerns were raised about the accuracy of the height estimate, suggesting that a height of 570 meters would result in a greater horizontal distance than observed. Additionally, the conversation touches on calculating the tension in a rope needed to move a 300 kg pallet of bricks, emphasizing the role of static friction.

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Daniiel
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I just want to double check that this is right, i have a feeling it's wrong because 570 metres doesn't seem right
1. Homework Statement
A hot air balloon is floating above the surface of a small lake and, in the early morning breeze, is drifting at 5km/h horizontally. A 4.5 kg bag is dropped from the balloon when it is directly above a boat. The bag lands in the water 15m away from the boat. How high is the balloon?


2. Homework Equations

x-x0 = (v0costheta)t
and the vertical formula y-y0 = (vosintheta)-1/2gt^2
3. The Attempt at a Solution
I used the horizontal motion formula to find t
i used the initial velocity as 1.39m/s
x=15
and got t=10.79
then used the vertical formula to find y
which was y = 4.9 x 116.43 = 570

I just followed an example of horizontal and vertical motion in my textbook
i think assuming the degree = 0 might of made my answer wrong or somthing
i kinda think that if the balloon was 570 the bag would of landed further than 15m
i apologize for my messiness

I also have another question, sorry I am not very confident with physics because there's always somthing I am not thinking about.

Question
In order to move a 300kg pallet of bricks across a smooth surface you need a rope tied to the back of your car. If the static friction is 0.2, how much tension is in the rope when the pallet just starts to move?

i just did 300 x 9.8 = 2940N
0.2 x 2940 = 588N
so is the required force to move the pallet 588+2940?
and the tension in the rope when it starts to move 588+2940?
 
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The first one is correct. People generally underestimate how quickly things fall. In reality the speed at the end of the fall would be 105 m/s and air friction would reduce this a lot.

To move the pallet you only have to overcome the static friction. The 2940 N is the weight of the pallet, you don't have to lift it.
 
Thanks a lot for your reply.
So, just to clarify.
If there was no static friction on a smooth surface, you wouldn't need any force to pull the pallet?
Like you don't have to over come the normal or anything?

also just to double check, if you only need to over come 588N would the tension in the rope when it begins to move be 588N?
Or like >588 so like it begins to move at 589
 
You still need a force to accelerate the pallet, but any force greater than 0 will do.
The acceleration will be very small for a small force (f = ma)
 

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