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basilrazi · Apr 14, 2009, 01:46 AM

A stone is dropped from a height of 51.2m above the ground.
How far does it fall during the last second of its motion?:confused:

Thanks

Unknown008 · Apr 14, 2009, 03:10 AM

Use $s = \frac{1}{2} gt^2$ where s = displacement, g = acceleration due to gravity and t = time. You'll then obtain the time of fall.

Remove one second from that time, and use

$v = u + at$ to find the speed v of the stone at that time.

Now use

$s = ut + \frac{1}{2}at^2$ where u = v from above, t = 1, a = acceleration due to gravity and s the distance you're looking for.

Hope that helps! :)

nurul azlina · Aug 9, 2009, 01:03 AM

A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard 3.2s later. If the speed of sound is 340m/s, how high is the cliff?

Unknown008 · Aug 9, 2009, 04:23 AM

You should have started another thread for you, nurul azlina.

Let the time taken for the stone to fall be $t_1$

And the time for the sound to move up be $t_2$

You have $t_1+t_2 = 3.2$

The distance the stone travels and that which the sound travels is the same.

So, let's call that distance 'd'.

Speed = Distance/Time, right?

Hence,

$340 m/s= \frac {d (m)}{t_2}$

and $t_2(s)= \frac {d (m)}{340 m/s}$

Now, see the formula for distance up there? $s=\frac12 gt^2$

The distance stone to falls is $d=\frac12 g(t_1)^2$

That means that $t_1 = \sqrt{\frac{2d}{g}}$

You know that g = 9.81 m/s^2

And replacing in the first equation, we have:

$t_1+t_2 = 3.2$

$t_1 = \sqrt{\frac{2d}{g}}$

$t_2(s)= \frac {d (m)}{340 m/s}$

So,

$\sqrt{\frac{2d(m)}{9.81(m/s^2)}}+\frac {d (m)}{340 (m/s)} = 3.2s$

And solve for 'd' to get the answer! :)