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

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! :)

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?

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! :)