Hi...

Q) A stone is thrown vertically upwards with a speed of 16m/s from a pointhmetres above the ground. The stone hits the ground 4s later. Find the value ofh.

Using suvat equations, how is this done?

Thanks alot

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- Jun 6th 2007, 01:58 PMApexFalling object
Hi...

Q) A stone is thrown vertically upwards with a speed of 16m/s from a point*h*metres above the ground. The stone hits the ground 4s later. Find the value of*h*.

Using suvat equations, how is this done?

Thanks alot - Jun 6th 2007, 02:19 PMJhevon
the SUVAT equation we need here is:

$\displaystyle s = vt - \frac {at^2}{2} = ut + \frac {at^2}{2}$

where $\displaystyle u$ is the initial velocity, $\displaystyle v$ is the final velocity, $\displaystyle s$ is the displacement or position, and $\displaystyle a$ is the acceleration.

However, since we did not start on the ground, $\displaystyle s$ has to be modified by some constant. that is, we will use

$\displaystyle s = ut + \frac {at^2}{2} + C$ where $\displaystyle C$ is the height we are when time, $\displaystyle t$, is zero.

now $\displaystyle u=16$, and $\displaystyle a = -9.8$

So, $\displaystyle s = 16t + \frac {-9.8t^2}{2} + C$

$\displaystyle \Rightarrow s = 16t - 4.9t^2 + C$

Now, after 4 seconds, the ball hits the ground, so it's height is zero.

$\displaystyle \Rightarrow s(4) = 0$

$\displaystyle \Rightarrow 0 = 16(4) - 4.9(4)^2 + C$

$\displaystyle \Rightarrow C = 14.4$

So, $\displaystyle s = -4.9t^2 + 16t + 14.4$

So $\displaystyle h = 14.4$, since that is what $\displaystyle s$ is when $\displaystyle t = 0$ - Jun 6th 2007, 03:24 PMApex
Thanks...I kept getting something like 13.06(my way). The solution in the paper returns the same answer as you, but it is done like this:

s=ut + 0.5at^2

and

s= ?

u= 16

v= -

a= -9.8

t= 4

What I don't understand, is why if it's done this way (using these values), it actually gives the value of h, as oppose to the distance*from*h, to where v=0, and then to the ground. i.e., the whole journey of the rock minus the distance h.

Can you please explain how this method works?

Thanks

Apex - Jun 6th 2007, 03:51 PMJhevon
think of it this way. we can use $\displaystyle h$ instead of $\displaystyle C$.

Now we know that $\displaystyle s = ut + 0.5at^2$. however, this equation assumes that the displacement is zero when time is zero. (if you plug in $\displaystyle t=0$ we get 0). here that is not the case. when time is zero, we are at a distance $\displaystyle h$ above the ground. so we must have:

$\displaystyle s = ut + 0.5at^2 + h$

Now if we plug in $\displaystyle t=0$, we get the distance $\displaystyle h$, which is what we want.

Now we simply must find $\displaystyle h$. we know that $\displaystyle s = 0$ when $\displaystyle t = 4$, so we simply plug those values in to find $\displaystyle h$

i'm not sure if i answered your question. did i? - Jun 6th 2007, 04:12 PMApex
Oh right, I see what you mean.

Thanks very much for your help:)

br

Apex