Projectile Problem: Freefall of Falcon at 200mi/h for 100m

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SUMMARY

The discussion revolves around calculating the vertical distance a falcon falls while traveling horizontally at a speed of 200 mi/h (converted to 89.4 m/s) over a distance of 100 meters. The correct vertical distance fallen is approximately 6.15 meters, derived from the time of fall calculated as 1.12 seconds using the formula for horizontal motion. The vertical distance is determined using the equation -4.9 m/s² * t², confirming that the falcon's horizontal speed remains constant while it falls. The calculations align with the principles of projectile motion, where the time of fall is equal to the time taken to cover the horizontal distance.

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  • Knowledge of unit conversions, specifically from miles per hour to meters per second
  • Ability to perform calculations involving significant figures
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of projectile motion and freefall dynamics.

gasapple
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What am I doing wrong here? I am trying to arrive at a correct answer of 6.13m but can't get there...

A falcon is moving horizontally at it's top speed (200 mi/h) at a height of 100m off the ground when it brings its wings in and drops into a freefall. How far does the bird fall vertically while traveling horizontally a distance of 100m?

I've converted the speed into 89.4 m/s (from 200 mi/h) and then isn't it simply a freefall problem for a duration of 100m? If someone could help me set it up so I can visualize it I would be appreciative. I guess I've got the initial x and y displacement relative to the Earth as well as the initial velocity, but now what?

Thanks in advance!
 
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find time taken for 100m horizontal, with constant velocity. What vertical distance it falls in that time? (0.5gt^2)
 
So would the constant velocity for that 100m segment simply be gravity or the x-component of the velocity (top speed)? I would guess that if you separate the components it would simply be freefall at gravity for 100m?

Thanks for your help...
 
the horisontal distance = Cos(angle)*V initial* time travelled.

angle is zeor so cos 0 = 1 so you have 100 = 89.4*t and you solve for time.

the vertical distance = Sin(angle)*V initial*t + 1/2 g*t^2

Sin of 0 degrees cancels the speed, so vertical fall would be -4.9m/s^2 * t^2 that you found in the first equation.

that's how i see it...but i only took phisics one in my junior year high school. i don't know the numeric outcome...give it a quick try.

edit: hold on i'll run the numbers for you.

t of fall is 1.12 seconds (correctly sig figured assuming the 100 has 3 sig figs.

-4.9*1.12^2= -6.15 if correctly sig figured or +6.15 if you're analizing a fall...

you get it...im insulting your intelligence here with the positive/negative thing.

my answer is slightly larger than the book's...by 2 centimeters becasue of repeated roundups due to sig figs rule. if i were to work with the 14 decimals i'd probably get 6.13
 
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OK, I'll take a look at it. Thanks for pondering it... I really appreciate it!
 
Think the bird falls for 1.12 seconds?
 
Ahh yes, you stated that in your response... THANKS! Now I know you're right - and I'm on the right track!
 
yea. what i did not write is how i came up with this i mean thinknig process.

well the bird is moving horisontally and suddenly starts to move vertically as well. horisontal speed won't change...and we only care about how far it falls during the time that it moves horisontally that distance. so the time of fall is the smae as the time it needs to travel that distance.

that's why the time is 1.12 in one equation and can be plugged into second. sorry for confusion...
 
Nah, makes perfect sense to me now. It's just a simple projectile - like a bullet fired from a gun with a dropped object - disregarding curvature of the earth, both would hit simultaneously because same y component. I think the time factor with this problem threw me off... thanks for explaining it...
 

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