How long does it take the ball to fall?

[Lunasly]

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Hello everyone,

Q. If a ball is dropped from a height of 10m at a constant acceleration of 9.8 m/s/s, then how long does it take to hit the ground.

I did this in my head and said that it should take 1 second to hit the ground, but that is not the case. Apparently, its 1.41 seconds. I used the formula Vf = Vit + 0.5at^2 and I got 1.41, but I am not understanding it conceptually.

We always say that if we launch a projectile with a Vy of 30 m/s then it will take 3 seconds to reach the apex because that is where Vy = 0. So that means every second the velocity as we move up is decreasing by a value of 10. Why is this same concept not true for the first example that I proposed?

Thanks,
Lunasly.

 

[milski]

Velocity is continuously increasing/decreasing. In one second the velocity will change from 0 to 10 m/s. But it will be 10 m/s only at the end of this second. During the rest of the period it will be something smaller, thus the distance at the end of the first second will be something smaller than 10 m.


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[Lunasly]

Ah thats why it doesn't take 1s exactly because during the second the velocity is smaller and not exactly at 10 m/s. Thus it taker longer then 1s (slightly) to travel 10m.

So in my second example: On the way up, we aren't dealing with any specific distance, but we know that at the apex the velocity is 0. Thus if we start at 20 m/s for Vy, then that means it will be at the end of exactly 2 seconds that our velocity becomes 0 (thus we are at the apex and the time of flight to the apex is 2 seconds).

Essentially, then, the two examples I gave are different in that I am provided with different information. In my first example, I am given a distance and asked how long it would take to travel that distance. In my second example. I am asked the same question (time to travel a certain distance – in this case the apex) but because I am not provided with a specific distance to the top, I can say that because my velocity at the apex is 0 m/s and I start with a vertical velocity of 20 m/s, then that means it is after 2 seconds that my velocity will be 0 and thus my time to the apex is 2 m/s.

Conclusion: I was asked the same question (time to reach something), but provided with different circumstances.

Can someone please confirm if I am correct here?

Thanks,
Lunasly.

 

[ridethecliche]

Ah thats why it doesn't take 1s exactly because during the second the velocity is smaller and not exactly at 10 m/s. Thus it taker longer then 1s (slightly) to travel 10m.

So in my second example: On the way up, we aren't dealing with any specific distance, but we know that at the apex the velocity is 0. Thus if we start at 20 m/s for Vy, then that means it will be at the end of exactly 2 seconds that our velocity becomes 0 (thus we are at the apex and the time of flight to the apex is 2 seconds).

Essentially, then, the two examples I gave are different in that I am provided with different information. In my first example, I am given a distance and asked how long it would take to travel that distance. In my second example. I am asked the same question (time to travel a certain distance – in this case the apex) but because I am not provided with a specific distance to the top, I can say that because my velocity at the apex is 0 m/s and I start with a vertical velocity of 20 m/s, then that means it is after 2 seconds that my velocity will be 0 and thus my time to the apex is 2 m/s.

Conclusion: I was asked the same question (time to reach something), but provided with different circumstances.

Can someone please confirm if I am correct here?

Thanks,
Lunasly.

Acceleration is constant and opposite to direction of upward travel.

vf=vi-at in your case

so:
0=20-10(t)
t=2

So now coming down the other way, it'll take 2 seconds to reach 20m/s. The distance, however, will be 20 meters.

vf^2=vi^2+2ad
20^2=0+20d
d=20

lets see for 30.
900=0+20d

And there the equality changes. So ~2secs is the 'equivalence' point. Don't use this as the only example.

In general, 2 is a bad number to use for effects since the square and multiplier effects (double vs square) can be confused.

Here's a helpful link: http://www.grc.nasa.gov/WWW/k-12/airplane/mofall.html

 

[ridethecliche]

Sassy answer:

Till a boy hits puberty.

 

[milski]

Ah thats why it doesn't take 1s exactly because during the second the velocity is smaller and not exactly at 10 m/s. Thus it taker longer then 1s (slightly) to travel 10m.

So in my second example: On the way up, we aren't dealing with any specific distance, but we know that at the apex the velocity is 0. Thus if we start at 20 m/s for Vy, then that means it will be at the end of exactly 2 seconds that our velocity becomes 0 (thus we are at the apex and the time of flight to the apex is 2 seconds).

Essentially, then, the two examples I gave are different in that I am provided with different information. In my first example, I am given a distance and asked how long it would take to travel that distance. In my second example. I am asked the same question (time to travel a certain distance – in this case the apex) but because I am not provided with a specific distance to the top, I can say that because my velocity at the apex is 0 m/s and I start with a vertical velocity of 20 m/s, then that means it is after 2 seconds that my velocity will be 0 and thus my time to the apex is 2 m/s.

Conclusion: I was asked the same question (time to reach something), but provided with different circumstances.

Can someone please confirm if I am correct here?

Thanks,
Lunasly.

Correct, you were provided initial and final displacement in one case and initial and final velocity in the other.


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[Izak]

I find the fastest method is to use the EK strategy. If it falls for 1 second, its final v will be 10 (since accelerate due to gravity g is 10m/s^2), thus the average is 10-0/2 = 5. The average v times t gives you displacement. So 5 times 1 = 5 m. Anything falling with g as acceleration and no air resistance, will fall 5 meters in 1 second.

Try 2 seconds. In 2 seconds, your final v will be 20, and your average t will be 20/2=10m/s. So 10m/s x 2 sec = 20 meters! For 10 meters, it would be 1.4 seconds.

Also, a cool formula is v=Radical(2gh). So using the initial v in the y direction, you can get the max height.

Hope you find this useful.

 

[MedPR]

h=1/2gt^2. Done.

 

[milski]

h=1/2gt^2. Done.

Half done, the other question was about v=g*t. Why is this thread still alive? I was looking forward to something fresher. 😎