Math Is Fun Forum

Daniel123

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Registered: 2007-05-23

Posts: 663

Equations of Motion

A stone is thrown vertically upwards with an initial velocity of of 29.4m/s from the top of a tower 34.3m high. Find:

a) the time taken to reach the maximum height

b) the total time which elapses before it reaches the ground

Quick response would be appreciated.

Thanks

JaneFairfax

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Registered: 2007-02-23

Posts: 6,868

Re: Equations of Motion

Take g = 9.8 ms[sup]−2[/sup], yeah?

(a) 3 s.

(b) 7 s

Quick enough?

Last edited by JaneFairfax (2007-09-11 07:40:46)

Daniel123

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Registered: 2007-05-23

Posts: 663

Re: Equations of Motion

haha wonderfully quick thanks... however, how did you do it? wouldn't you need to know the acceleration of the stone upwards?

JaneFairfax

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Registered: 2007-02-23

Posts: 6,868

Re: Equations of Motion

Sorry, my second answer was wrong – just edited it.

JaneFairfax

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Registered: 2007-02-23

Posts: 6,868

Re: Equations of Motion

Actually you need to take g = −9.8 ms[sup]−2[/sup] for (a) if you are taking upward as +ve. Then use v = u + at with u = 29.4 ms[sup]−1[/sup], v = 0 ms[sup]−1[/sup], a = −9.8 ms[sup]−2[/sup].

The ball will spend 3 seconds going upwards, and another 3 seconds coming down before commencing its freefall down the building. At the start of its fall down the building, its velocity is 29.4 ms[sup]−1[/sup] downwards.

For (b) it will be easier to take downward as +ve. Use s = ut + ½at[sup]2[/sup], where s = 34.3 m, u = 29.4 ms[sup]−1[/sup], a = 9.8 ms[sup]−2[/sup], to find the time it takes to fall down the building.

It only takes 1 second to do it.

Hence the total time the object spends in the air is 7 seconds.

Last edited by JaneFairfax (2007-09-11 07:54:48)

Daniel123

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Registered: 2007-05-23

Posts: 663

Re: Equations of Motion

Ok thankyou, but I don't understand why the acceleration of the stone is 9.8m/s²?

luca-deltodesco

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Registered: 2006-05-05

Posts: 1,470

Re: Equations of Motion

goddamn it you beat me to it (all deleted)

acceleration of stone is 9.8m/s^2 because that is the acceleration due to gravity, which is the only acceleration in effect

and its negative 9.8 since you are taking positive to be up

Last edited by luca-deltodesco (2007-09-11 07:58:14)

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Daniel123

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Registered: 2007-05-23

Posts: 663

Re: Equations of Motion

Hmm I'm getting confused.

The way I did it was:

a) time for stone to reach maximum height

b) total time stone is in air

to calculate final velocity when stone hits the ground:

to calculate time stone takes to reach ground:

So 3s to reach maximum height + 3s to reach level with building + 1s to hit ground = 7s.

But.... I don't understand why s = ut + ½t² gives you 7s. If you are using 29.4m/s as the initial velocity, and 34.3m as s, then wouldn't the result give you the time it takes for the stone to hit the ground from when it is level with the building? Why does it give 7s, and not 1s?

Long day.

Thanks

Last edited by Daniel123 (2007-09-11 08:33:01)

luca-deltodesco

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Registered: 2006-05-05

Posts: 1,470

Re: Equations of Motion

yes it does, but remember, its level with the building at 2 occasions, when you first throw it, and when it has came back down, and since you are modelling it with its initial velocity, it models the full projectory up, then back down all the way

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krassi_holmz

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Registered: 2005-12-02

Posts: 1,905

Re: Equations of Motion

I'm trying learn to apply Neuton rules directly, and I've solved it this way:

S is maximal, then S' is 0, so:

When the rock goes back to earth, S(t) = 0, solving:

Last edited by krassi_holmz (2007-09-12 09:41:33)

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Daniel123

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Registered: 2007-05-23

Posts: 663

Re: Equations of Motion

Don't look at me... I haven't even done integration yet...