# Acceleration and deceleration due to gravity?

• May 13th 2011, 08:04 AM
jebus197
Acceleration and deceleration due to gravity?
If an arrow on Earth is fired into the air and reaches a hight of 125 m and penetrates the ground where it lands to a depth of 25 cm, what is it's speed when it hits the ground and what is it's deceleration? Of course the acceleration due to gravity is 9.8 m s$\displaystyle ^-2$, but I'm stumped on how to calculate it's final speed. There must be a quck and easy way to do this surely?
• May 13th 2011, 09:01 AM
TheEmptySet
Quote:

Originally Posted by jebus197
If an arrow on Earth is fired into the air and reaches a hight of 125 m and penetrates the ground where it lands to a depth of 25 cm, what is it's speed when it hits the ground and what is it's deceleration? Of course the acceleration due to gravity is 9.8 m s$\displaystyle ^-2$, but I'm stumped on how to calculate it's final speed. There must be a quck and easy way to do this surely?

We don't care about the arrow going up. At its highest point we know that its velocity is 0 and its height is 125m. Using the kinematic equations gives

$\displaystyle x_0=125 \quad x_f=0 \quad v_0=0 \quad v_f=? \quad a=-g$

so the equation is

$\displaystyle 0-125=0\cdot t -\frac{g}{2}t^2 \iff t^2=\frac{250}{g} \implies t=\frac{5\sqrt{10g}}{g}$

This gives the time when the arrow hits the ground now use the equation

$\displaystyle v_f=v_0+at=0-gt \quad v_f=-5\sqrt{10g}$

So this is the velocity when the arrow hits the ground.
• May 13th 2011, 09:35 AM
jebus197
That read a bit like Greek to me, sorry. I'm a total beginner. The equation (I think!) I have to use are mg $\displaystyle \Delta$ h = 1/2 mv^2 and I have to rearrange this to obtain the velocity.

I get that you can divide both sides by m to discount the mass. So this gives g $\displaystyle \Delta$ h = 1/2 v^2

But since I want to make v the subject I can rearrange again to give:

1/2 v^2 = g $\displaystyle \Delta$ h

But how do I get rid of the half on the left side of the equation? I presume I have to take the square root of both sides of the equation, so that I can just obtain v, but that's where I get stuck!

Your help is much appreciated!
• May 13th 2011, 09:42 AM
TheEmptySet
Quote:

Originally Posted by jebus197
That read a bit like Greek to me, sorry. I'm a total beginner. The equation (I think!) I have to use are mg $\displaystyle \Delta$ h = 1/2 mv^2 and I have to rearrange this to obtain the velocity.

I get that you can divide both sides by m to discount the mass. So this gives g $\displaystyle \Delta$ h = 1/2 v^2

But since I want to make v the subject I can rearrange again to give:

1/2 v^2 = g $\displaystyle \Delta$ h

But how do I get rid of the half on the left side of the equation? I presume I have to take the square root of both sides of the equation, so that I can just obtain v, but that's where I get stuck!

Your help is much appreciated!

Note that you will get the same answer I did but you are using the conservation of energy

$\displaystyle mg \Delta h=\frac{1}{2}mv^2 \iff 2\Delta h =v^2 \iff v=\sqrt{2g\Delta h}$

So the change in height is 125

$\displaystyle v=\sqrt{2g(125)}=\sqrt{250g}=\5\sqrt{10g}$
• May 13th 2011, 10:21 AM
jebus197
Let me get this right by saying it in words. The very last line of your equation reads

v = the square root of 2 x g (x 125).

So v = the square root of 250 x g

So 250 x 9.8 m s-^2 = 2450 m s^2

So the square root of 2450 m s^2 = 49.5 m s^-1

So v= 49.5 m s^-1 ???

I really very sorry for being so dumb. I'm an adult learner and until a few months ago I barely even knew any maths at all.
• May 13th 2011, 10:28 AM
TheEmptySet
Yes that is what I get.
• May 13th 2011, 10:54 AM
jebus197
Thanks man. It's a big ask. But I've been scrambling around the net looking and just getting more and more confused. But what about the deceleration? I'm not asking for the answer, I'm just asking for a simple method of calculating it. If the arrow penetrates the ground to a depth of 0.25 m before coming to a stop and it's velocity is 49.5 m s^-1, then what is the rate of deceleration? Is this simply a division of 0.25 m / 49.5 m s^-1? Clearly this is a nonsense answer as the m/m would cancel each other out?
• May 13th 2011, 11:32 AM
topsquark
Quote:

Originally Posted by jebus197
Thanks man. It's a big ask. But I've been scrambling around the net looking and just getting more and more confused. But what about the deceleration? I'm not asking for the answer, I'm just asking for a simple method of calculating it. If the arrow penetrates the ground to a depth of 0.25 m before coming to a stop and it's velocity is 49.5 m s^-1, then what is the rate of deceleration? Is this simply a division of 0.25 m / 49.5 m s^-1? Clearly this is a nonsense answer as the m/m would cancel each other out?

Look at it as a 1-D motion problem. You have the initial speed (49.5 m/s), you know what the final speed is (0 m/s), and you know how far it went (0.45 m). So you've got variables y, v, v0 (and y0 = 0 m by the usual coordinate system), and you are looking for a. Sounds like a good case for
$\displaystyle v^2 = v_0^2 + 2a(y - y_0)$

-Dan
• May 13th 2011, 11:43 AM
jebus197
I haven't got a clue what that means. Please appreciate that until 2 or 3 months ago you are talking to someone who was practically maths illiterate. Isn't there just a simple way to calculate deceleration using numbers and/or words?
• May 13th 2011, 11:44 AM
jebus197
How about a very similar numerical example?
• May 13th 2011, 07:07 PM
Soroban
Hello, jebus197!

Here is another explanation.
I hope you can following the reasoning.

Quote:

If an arrow on Earth is fired into the air and reaches a height of 125 m
and penetrates the ground where it lands to a depth of 25 cm,
(a) what is its speed when it hits the ground?
(b) What is its deceleration?

Using "meters", the height function is: .$\displaystyle y \:=\:h_o + v_ot - 4.9t^2$

. . where: .$\displaystyle \begin{Bmatrix}h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \end{Bmatrix}$

Assuming the arrow is shot from ground level, we have: $\displaystyle h_o = 0.$

. . We have: .$\displaystyle y \:=\:v_ot - 4.9t^2$

When does the arrow strike the ground?
. . That is, when does $\displaystyle y = 0$?

We have: .$\displaystyle v_ot -4.9t^2 \:=\:0 \quad\Rightarrow\quad t(v_o-4.9t) \:=\:0$

. . Hence: .$\displaystyle t = 0,\;t = \frac{v_o}{4.9}$

The arrow is at ground level at $\displaystyle t = 0$ (when the arrow is fired into the air),
. . and $\displaystyle \frac{v_o}{4.9}$ seconds later (when it strikes the ground).

Fact: The arrow spends half its time going up and the other half going down.

So, the arrow reaches its peak at: .$\displaystyle t \:=\:\frac{1}{2}\left(\frac{v_o}{4.9}\right) \:=\:\frac{v_o}{9.8}$ seconds.

Hence, the maximum height is: .$\displaystyle y \;=\;v_o\left(\frac{v_o}{9.8}\right) - 4.9\left(\frac{v_o}{9.8}\right)^2 \;=\;\frac{v_o^2}{19.6}$

$\displaystyle \text{The maximum height is 125 meters: }\;\frac{v_o^2}{19.6} \:=\:125$

. . $\displaystyle v_o^2 \:=\:2450 \quad\Rightarrow\quad v_o \:=\:\sqrt{2450} \:=\:35\sqrt{2}$

The initial velocity is: .$\displaystyle 35\sqrt{2}\text{ m/sec}$ upward.

Fact: The speed at which the arrow strikes the ground
. . . . . is equal to its initial speed.

$\displaystyle \text{(a) Therefore, the arrow hits the ground at }35\sqrt{2}\text{ m/sec.}$

• May 14th 2011, 03:44 AM
topsquark
Quote:

Originally Posted by jebus197
I haven't got a clue what that means. Please appreciate that until 2 or 3 months ago you are talking to someone who was practically maths illiterate. Isn't there just a simple way to calculate deceleration using numbers and/or words?

Aside from solving simultaneous equations there is no other way to approach this part of the problem. I appreciate your difficulty, but to tackle a question like this you need to learn the basic Physics behind it. You can't put the cart before the horse. The Math here isn't that hard once you get a little experience with it. I'd suggest looking at some simpler problems and work your way up.

-Dan