# Falling object

• Jun 6th 2007, 01:58 PM
Apex
Falling 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 PM
Jhevon
Quote:

Originally Posted by Apex
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

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 PM
Apex
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 PM
Jhevon
Quote:

Originally Posted by Apex
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

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 $\displaystyle h$