You are not logged in.
Pages: 1
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
Offline
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)
Offline
haha wonderfully quick thanks... however, how did you do it? wouldn't you need to know the acceleration of the stone upwards?
Offline
Sorry, my second answer was wrong just edited it.
Offline
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)
Offline
Ok thankyou, but I don't understand why the acceleration of the stone is 9.8m/s²?
Offline
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)
The Beginning Of All Things To End.
The End Of All Things To Come.
Offline
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)
Offline
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
The Beginning Of All Things To End.
The End Of All Things To Come.
Offline
I'm trying learn to apply Neuton rules directly, and I've solved it this way:
Last edited by krassi_holmz (2007-09-12 09:41:33)
IPBLE: Increasing Performance By Lowering Expectations.
Offline
Don't look at me... I haven't even done integration yet...
Offline
Pages: 1