How long does it take to free fall from Mt. Everest?

[HamzaPhysical]

1.If i were to jump of of Mt. Everest (29,090 ft), how long will it take me to hit the ground? The initial velocity is 0, the pull of gravity is equal to 9.81m/s^2(acceleration).

I got 42 seconds, but i didnt calculate terminal velocity so I am pretty sure my answer is wrong

to calculate time i used the equation t= the square root of twice the distance divided by the square root of the acceleration.
I converted feet into meters by multiplying it by 30 cm/ft.
from then on i just plugged the info in.

 

[phinds]

How are we supposed to check your work if you don't SHOW you work?

 

[gneill]

1.If i were to jump of of Mt. Everest (29,090 ft), how long will it take me to hit the ground? The initial velocity is 0, the pull of gravity is equal to 9.81m/s^2.

I got 48 second, but i didnt calculate terminal velocity so I am pretty sure my answer is wrong

Hi HamzaPhysical, Welcome to Physics Forums.

In future, please use the Posting Template provided when you start a thread in the homework areas. This is a forum requirement.

Also, can you show your attempt? How did you arrive at 48 seconds?

 

[lightbender]

Well, First did you convert 29090 ft to meters?

 

[HamzaPhysical]

yes

 

[lightbender]

Then, did you use one of the four kinematic equations?

 

[HamzaPhysical]

i derived the time equation that you see from one of them

 

[PhanthomJay]

i derived the time equation that you see from one of them

Looks like you corrected your earlier error and arrived at the correct answer in the absence of air resistance. No need to convert feet to meters if you use the acceleration of gravity as 32.2 feet per second per second (often rounded to 32). Now if you consider air resistance, the problem becomes more complex, and the time of fall increases substantially because terminal velocity will be reached within a few seconds, the value of which depends on whether you fall belly up or head first , amongst other factors. The problem didn't mention air resistance did it, or are you just curious?

 

[Matterwave]

I'm guessing the problem is not being very realistic, it just wants you to find the time of free fall for 29,000 feet. I mean, Mount Everest might be ~29,000 feet above sea level, but there's no way you will reach the sea from free-falling from the peak of Mount Everest. Realistically, you will hit the side of the mountain after not more than a few seconds and then tumble your way down to a relatively level plateau (or just get caught in the ice and freeze to death).

 

[Ravnoor]

I'm guessing the problem is not being very realistic, it just wants you to find the time of free fall for 29,000 feet. I mean, Mount Everest might be ~29,000 feet above sea level, but there's no way you will reach the sea from free-falling from the peak of Mount Everest. Realistically, you will hit the side of the mountain after not more than a few seconds and then tumble your way down to a relatively level plateau (or just get caught in the ice and freeze to death).

1.If i were to jump of of Mt. Everest (29,090 ft), how long will it take me to hit the ground? The initial velocity is 0, the pull of gravity is equal to 9.81m/s^2(acceleration).

I got 42 seconds, but i didnt calculate terminal velocity so I am pretty sure my answer is wrong

to calculate time i used the equation t= the square root of twice the distance divided by the square root of the acceleration.
I converted feet into meters by multiplying it by 30 cm/ft.
from then on i just plugged the info in.

 

[CWatters]

Your answer of 42 seconds appears correct if you ignore air resistance. To get 2min 43 sec they appear to have used the terminal velocity of a human which is about 52m/s according to Wikipedia.