Fire an arrow at a moose as it runs away

[Girn261]

http://imgur.com/a/FKkHL

1. Homework Statement
A moose is running away at 25m/s. You fire an arrow at 30 degrees when it's 120m away. What must the velocity of the arrow be in order to hit it?

Homework Equations

The Attempt at a Solution

Please see attached picture, I'd type it out but I'm unsure how some of you type it out in the box which allows for several different math notations etc and so forth.

I assume it's because the 120m in the quadratic equation should be a postive and not a negative? But the distance is increasing as the moose runs away

 

[Eclair_de_XII]

But the distance is increasing as the moose runs away

Exactly. For starters, what you want to do is express your range $R$ as the sum of the initial distance and the distance the moose has traveled from that point at time $t$.

 

[Girn261]

Exactly. For starters, what you want to do is express your range $R$ as the sum of the initial distance and the distance the moose has traveled from that point at time $t$.

I did put 120+x=2.548u , hmm. Anymore tips?

 

[Eclair_de_XII]

What is $x$? Express it as a function of $t$.

 

[Girn261]

What is $x$? Express it as a function of $t$.

Hmm X is S (distance). Now I'm so confused haha. This problem was tough for me.

 

[Eclair_de_XII]

$x$ is the distance traveled by the moose running at speed $25\frac{m}{s}$. What distance does it travel after four seconds of running away?

 

[Girn261]

$x$ is the distance traveled by the moose running at speed $25\frac{m}{s}$. What distance does it travel after four seconds of running away?

100m , S=vt. But both S and t are unknowns in the equation. Hmmm

 

[Eclair_de_XII]

$S=vt$

Okay, let's say then that $S$ is the distance that the moose runs away. It's good that you realize that $S(t)=vt$. It's even better that you know that you have to find $t$. It's time to look at your equations of motion. This will likely take a while, by the way. First off, let's create a new time variable $t_0$--the time it takes for your arrow to reach its maximum height in its trajectory. You want to take the $y$ component of your $u$ velocity and calculate how long it takes to fall to zero. Can you do that?

 

[Girn261]

Okay, let's say then that $S$ is the distance that the moose runs away. It's good that you realize that $S(t)=vt$. It's even better that you know that you have to find $t$. It's time to look at your equations of motion. This will likely take a while, by the way. First off, let's create a new time variable $t_0$--the time it takes for your arrow to reach its maximum height in its trajectory. You want to take the $y$ component of your $u$ velocity and calculate how long it takes to fall to zero. Can you do that?

I'll try this when I get home. Thank you. So I guess the way I did it was incorrect. Finding the units to plug into the quadratic equation. I sort of followed another question that was answered the same way, but the "moose" was running towards the person instead of away from it.

 

[Girn261]

Okay, let's say then that $S$ is the distance that the moose runs away. It's good that you realize that $S(t)=vt$. It's even better that you know that you have to find $t$. It's time to look at your equations of motion. This will likely take a while, by the way. First off, let's create a new time variable $t_0$--the time it takes for your arrow to reach its maximum height in its trajectory. You want to take the $y$ component of your $u$ velocity and calculate how long it takes to fall to zero. Can you do that?

Tan30=y/120m Ym=69.28 , sin30=y/25m/s Yv = 12.5
So S=vt, 69.28m=12.5t. (12.5 being y velocity component, same for S) and t is 5.54s

 

[Eclair_de_XII]

What I was looking for was an expression of $t$ as a function of $v$, actually.

Anyway, tell me how to find the final velocity of an object given the initial velocity, acceleration, and time.

 

[Girn261]

What I was looking for was an expression of $t$ as a function of $v$, actually.

Anyway, tell me how to find the final velocity of an object given the initial velocity, acceleration, and time.

Vf=vi+at , hmm sorry, v=s/t

 

[Eclair_de_XII]

$v_f=v_i+at$

No, that's correct. Now, given a final velocity, an initial velocity, and acceleration, can you find $t$?

 

[Girn261]

No, that's correct. Now, given a final velocity, an initial velocity, and acceleration, can you find $t$?

Yes you can, so I assume for intial velocity you use 25m/s and final velocity u use 12.5m/s I found earlier. And acceleration being gravity. You get time being 1.274. I have a feeling this is wrong. Confusing

 

[Eclair_de_XII]

I'm sorry. I wasn't clear. I was talking about the launch velocity of the arrow $u$. You're not supposed to get a numerical value for $t$ yet.

 

[Eclair_de_XII]

@Girn261

To reiterate, you want to find the time that the y-component of $u$ (the velocity of the arrow) reaches zero. So express $t$ in terms of $g$ and $u_y$, first.

 

[ehild]

http://imgur.com/a/FKkHL

1. Homework Statement
A moose is running away at 25m/s. You fire an arrow at 30 degrees when it's 120m away. What must the velocity of the arrow be in order to hit it?

Homework Equations

The Attempt at a Solution

Please see attached picture, I'd type it out but I'm unsure how some of you type it out in the box which allows for several different math notations etc and so forth.

I assume it's because the 120m in the quadratic equation should be a positive and not a negative? But the distance is increasing as the moose runs away

I do not see your picture, so I show the problem in the attached one.

The mouse moves horizontally, the arrow performs projectile motion. x(t) is the distance from the point the arrow was shot at time t=0, y(t) is the height of the arrow. The mouse was at xo=120 m at t=0, moving with velocity of 25 m/s away, so its displacement is s=25t. You have to write out x(t) both for the mouse and the arrow, and also y(t) for the arrow. When the arrow hits the mouse, both x and y most be the same for them. As the mouse moves on the ground, y(t)=0 for the arrow at that moment.

 

[Girn261]

With the same formula as earlier, rearranged with time as function with u as 12.5, v as 25 I get 1.27s for time

I'm sorry. I wasn't clear. I was talking about the launch velocity of the arrow $u$. You're not supposed to get a numerical value for $t$ yet.

 

[Eclair_de_XII]

Okay, let me just list the variables for you.

$u_y$ is the y-component of your initial launch velocity
$g$ is acceleration due to gravity which acts on $u_y$
Let $u_y=v_i$ and $a=-g$.

Can you tell me what $v_f$ is supposed to be in $v_f=v_i+at$ if you want the time at which the arrow reaches maximum height?

 

[ehild]

Okay, let me just list the variables for you.

$u_y$ is the y-component of your initial launch velocity
$g$ is acceleration due to gravity which acts on $u_y$
Let $u_y=v_i$ and $a=-g$.

Can you tell me what $v_f$ is supposed to be in $v_f=v_i+at$ if you want the time at which the arrow reaches maximum height?

You do not need to work with the maximum height. That time is needed, when the arrow hits the ground as the mouse moves on the ground.

 

[Girn261]

Okay, let me just list the variables for you.

$u_y$ is the y-component of your initial launch velocity
$g$ is acceleration due to gravity which acts on $u_y$
Let $u_y=v_i$ and $a=-g$.

Can you tell me what $v_f$ is supposed to be in $v_f=v_i+at$ if you want the time at which the arrow reaches maximum height?

You do not need to work with the maximum height. That time is needed, when the arrow hits the ground as the mouse moves on the ground.

How's this look? I watched a lesson on something similar, except the moose was running towards.

 

[ehild]

View attachment 209961

How's this look? I watched a lesson on something similar, except the moose was running towards.

It looks grey on grey and I can read it. Why you did not type it in?
According to the last line, the result might be correct.