# How long does it take the penny to reach the ground?

## Homework Statement

A girl is on a ride at an amusement park. She is 30m above the ground and is ascending vertically at 5.0 m/s. While she is ascending, she holds her arm out and releases a penny. How long does it take th penny to reach the ground?

## Homework Equations

y=yi+vit+0.5(at^2)

## The Attempt at a Solution

vi=5.0 m/s
yi=30 m
a=-9.8 m/s^2 (Letting up be positive.)
y=0

0=30m+50m/s t+0.5(-9.8 ,/s^2) t^2
For the quadriatic formula, C= 30 m, B=50 m/s, and A= -4.9 m/s^2

After using the formula the positive answer is t≈10.79 s. Is this correct? Thanks for a reply!

Chestermiller
Mentor

## Homework Statement

A girl is on a ride at an amusement park. She is 30m above the ground and is ascending vertically at 5.0 m/s. While she is ascending, she holds her arm out and releases a penny. How long does it take th penny to reach the ground?

## Homework Equations

y=yi+vit+0.5(at^2)

## The Attempt at a Solution

vi=5.0 m/s
yi=30 m
a=-9.8 m/s^2 (Letting up be positive.)
y=0

0=30m+50m/s t+0.5(-9.8 ,/s^2) t^2
For the quadriatic formula, C= 30 m, B=50 m/s, and A= -4.9 m/s^2

After using the formula the positive answer is t≈10.79 s. Is this correct? Thanks for a reply!
Try again. B should be 5.0, not 50.

nasu
Gold Member
The initial speed is 5 m/s and not 50 m/s, right?
With 5 m/s the time is about 3s.

Thanks to both of you!

I can't believe I forgot a decimal and thought it was right for the entire million-step problem. I thought the answer was a bit too high. (P.S. @ nasu- Someone is most likely going to come snap at you for giving me a partial answer, but it's not like I'm not reworking the problem anyway. Just a heads up. )

After doing the exact same thing but remembering the decimal: T is approximately 3.04 s.

nasu
Gold Member

Ah, gotcha. Sorry, I know, I just had somebody else do the same thing and get snapped at, so I wasn't sure. Thanks for all your help!! :-)

Also, that is a really great estimate.

nasu
Gold Member
You can do it yourself. The numbers were nice here. And I took g=10 m/s^2.
So it goes up t1=0.5 s (the speed decreases by 10 m/s every second).
The up distance is $5t_1^2$ or $5\times (1/2)^2 =5\times (1/4)=1.25 m$
(5 is g/2 where g is 10 m/s^2; 1/2 is t1=0.5s)

Then it falls freely from 31.5 m. How long it takes?
$t_2^2=31.5/5= 6.25 s^2$
And this is the square of 2.5. So t2=2.5s and t1=0.5 s, total 3s.
This is the reason for the estimate to be so close. If it were some not so nice a number (a square of something familiar), I'll have to estimate the square root and it will be less accurate.

Doing it by pieces may help you to understand better. But is just a suggestion. If it's confusing, just disregard it. I only wanted to show you that it was indeed an estimate, that it can be done without even writing something down.

That's really neat! Thanks!