Easy question on exam that I couldnt figure out HELP

[Aristodol]

A boy throws a rock horizontally 12 m/s at a pigeon on a fence 5m away. Unfortunatley the rock the missed. By how much did the rock miss?

I ve got my kinematic stuff down pat, but I just had no idea how to go about this problem.

thanks for any help.

 

[Doc Al]

Assume that the boy aimed directly at the pigeon.

 

[berkeman]

...and how long did it take the rock to hit the fence?

 

[Aristodol]

the time was not given, that why I had trouble with it. All the information above is exactly what it said on the test.

 

[Doc Al]

the time was not given...

You should be able to figure out the time it takes for the rock to hit the fence--you are given all the information needed.

 

[f(x)]

assuming g=10,i get it as ~0.5 m...

 

[arunbg]

Assuming the boy has sufficient height ie, the stone strikes the fence , all the information required is given in the question.
However your answer seems to be incorrect. Post your attempt if you can and we'll try to correct you.

 

[Doc Al]

assuming g=10,i get it as ~0.5 m...

Please give Aristodol the opportunity to solve his own problem. (And if you wish to help, don't just toss out answers.)

 

[Aristodol]

I no that assuming that x is 5 you get a time .42s i believe. but that's not the time we actually need so that is where i got stuck. I sat there for a long time pondering this question.

 

[Doc Al]

That is most definitely the time you need. Keep going.

 

[Aristodol]

Ive been trying to solve this problem for severall days I have been unable to do it. If you could please just show me the way to go about it it would be appreciated. I've tried many of time but i keep getting stuck.

 

[Doc Al]

Explain what you've been doing and where you are getting stuck. You have correctly calculated the time it takes for the rock to reach the fence.

Hint: If there were no gravity, the rock would go in a straight line and hit the pigeon. But gravity acts. During the time it takes the rock to get to the fence, how far does it fall?

 

[Aristodol]

I have established that t=.42s we no that Vx=Vox=12m/s also Vfy=0m/s ay=-9.8m/s O i see it now

Vf=Vo + at---> 0=Voy + -9.8(.42) Voy= 4.12m/s
now what's throwing me off is that i have the velocities for x and y and I am given a distance in 5 but that's not the correct distance

 

[Doc Al]

I have established that t=.42s we no that Vx=Vox=12m/s also Vfy=0m/s ay=-9.8m/s O i see it now

Good. You have the correct initial velocity, both horizontal and vertical components.

Vf=Vo + at---> 0=Voy + -9.8(.42) Voy= 4.12m/s

You have found the vertical component of the rock's speed at the moment it hits the fence. But what you need is the vertical distance it travels in that time.

now what's throwing me off is that i have the velocities for x and y and I am given a distance in 5 but that's not the correct distance

5 m is the horizontal distance it travels; you've already used it to find the time of travel.

What kinematic equation relates distance to time for constant acceleration?

 

[Aristodol]

y=4.12(.42) + 1/2 (-9.8)(.42)^2
1.73 + -.86
y=.87m ok?

 

[Doc Al]

y=4.12(.42) + 1/2 (-9.8)(.42)^2
1.73 + -.86
y=.87m ok?

Not exactly. The equation you are using (for vertical displacement) is this:
y = v_0 t + (1/2) a t^2

But what is the vertical component of the initial velocity?

 

[Aristodol]

huh? not exactly sure what u mean by that?

 

[berkeman]

huh? not exactly sure what u mean by that?

In an earlier post of yours, you correctly identified the initial vertical velocity of the rock. Doc is asking where you came up with the 4.12 term. What is it? It certainly is not the initial vertical velocity of the rock.

BTW, I think that the final answer you will get is for how far below the bird the rock hits, correct?

 

[Aristodol]

4.12 is indeed the initial vertical velocity I solved for it and he agreed that that was indeed the correct number.

Vf=Vo + at---> 0=Voy + -9.8(.42) Voy= 4.12m/s

 

[Doc Al]

4.12 is indeed the initial vertical velocity I solved for it and he agreed that that was indeed the correct number.

Nope. 4.12 m/s is not the correct initial vertical speed! 4.12 m/s downward is the vertical component of the velocity when the rock hits the fence--which is not needed to solve this problem.

Vf=Vo + at---> 0=Voy + -9.8(.42) Voy= 4.12m/s

I probably misread this before. For some reason, you have set Vf = 0. Why? (I had assumed that you were calculating Vf, not Vo. My bad. )

In any case, the initial vertical speed of the rock is given, no need to calculate it. Remember: It is thrown horizontally.