Projectile Question

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  • #1
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Homework Statement


An arrow fired horizontally at 41m/s travels 23m horizontally before it hits the ground. From what height was it fired?


Homework Equations


constant acceleration formula


The Attempt at a Solution


This is what i have done, however it is incorrect and the height of y is 1.54m(book's answers)
So to find y i had to find the time.
So...

x=23m
v-final(x)=41m/s
t=?
v-initial(x)=0m/s

using the constant acceleration formula:

[tex]x= \frac{v-initial - v-final}{2} * t[/tex]

[tex]23 = \frac{41}{2}t[/tex]

[tex]t= \frac{23}{20.5}[/tex]

t=1.12s


Now to find y

v-initial(y)=0m/s
a=-9.80m/s2
t=1.12s
y=?

i used y=v-initial*t + 0.5*a*t

eventually i get 6.15m
 

Answers and Replies

  • #2
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It is fired horizontally at 41m/s.

In the x-direction, using:

[tex] s = ut + \frac{1}{2}at^2 [/tex]

We discover that:

[tex] \frac{23}{41} = t [/tex]

As there is no acceleration horizontally

Then for the y-direction

[tex] s =ut+\frac{1}{2}at^2 [/tex]

So,

[tex] s=\frac{1}{2}g\left(\frac{23}{41}\right)^2 [/tex]

[tex] s = 1.54m [/tex]
 
  • #3
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It is fired horizontally at 41m/s.

In the x-direction, using:

[tex] s = ut + \frac{1}{2}at^2 [/tex]

We discover that:

[tex] \frac{23}{41} = t [/tex]
i don't understand how you have got [tex]\frac{23}{41}[/tex] from [tex] s = ut + \frac{1}{2}at^2 [/tex]. It seems like you have used the [tex]v=\frac{s}{t}[/tex] rule instead?

***Edit don't worry i understand after reading a=0
 
  • #4
The arrow moves in the x-axis direction without any gravity or any force affecting it's motion.

So, the kinematic equations can be used in each axis seperately.

Using the generic equation of motion (the only important one from which all others are easily derived);
[tex] x \ = \ x_0 \ + \ (v_0_x)t \ + \ \frac{1}{2} a t^2 [/tex]

we see that the only active variables are [tex] x = (v_0)t [/tex]

If you think about it, the arrow would move forever in this direction unless some force or collision stopped it.

Now, the problem has told us it doesn't move forever in the x-axis direction.
The problem said it moved 23m with an initial horizontal velocity of 41 m/s.
So, we know here that we are missing the time. We'll solve for the time.

[tex] t \ = \ \frac{x}{v_0} [/tex]
[tex] t \ = \ \frac{23m}{41m/s} [/tex]
[tex] t \ = \ 0.56s [/tex]

Using our knowledge about motion (and in particular the motion equation above) we realise that in the y-axis there is a force, gravity, pulling the arrow downwards. Lets look at the equation of motion for the y-axis

[tex] y \ = \ y_0 \ + \ (v_0_y)t \ - \ \frac{1}{2} a t^2 [/tex]

Note, in this version of the equation I've added a minus. I did this so that I can make the acceleration due to gravity as + 9.8m/s^2, you'd want to study this fact yourself to be sure you understand what I've done.

We realise that, as the arrow was shot straight (horizontally) so there is no initial velocity in this direction of motion.
From the question, we seek to find y_0 as this represents the initial height.
We also know the time that the arrow is in the air now.
Furthermore, we can set y = 0 because that will be the final height after the elapsed time.
The arrow will be at a height of zero because it has hit the ground & stopped moving.
We now have;

[tex] 0 \ = \ y_0 \ - \ \frac{1}{2} a t^2 [/tex]
[tex] y_0 \ = \ \frac{1}{2} a t^2 [/tex]
[tex] y_0 \ = \ \frac{1}{2} (9.8m/s^2) (0.56s)^2 [/tex]
[tex] y_0 \ = \ 1.53654 [/tex]
[tex] y_0 \ \cong \ 1.54 [/tex]

This is a foolproof method to follow. All you need to do is rely on the good ol' equation above & practice deriving all of the other equations of motion from this. Note, this equation itself is derived from constant acceleration & integral calculus.


EDIT-------------------------------

***Edit don't worry i understand after reading a=0
lol I took so long writing that that you understood it. Well here you go anyway ;)
 

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