Object Thrown from building help

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Homework Help Overview

The problem involves a stone being thrown from a height of 45 meters, landing 64 meters away. Participants are exploring various aspects of projectile motion, including time of flight, initial speed, and the velocity vector at impact.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating the time of flight and initial speed using kinematic equations. There are inquiries about the vertical and horizontal components of velocity at impact and how they relate to the overall motion.

Discussion Status

Some participants have provided insights into the independence of horizontal and vertical motion, while others are attempting to calculate the components of velocity at impact. There is an ongoing exploration of the angle of impact and its calculation.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available for discussion and the methods they can employ.

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Homework Statement


A boy throws a stone from the top of the building that is 45 m high. It lands 64 m away. Find the following:

a) time of flight
b) initial speed
c) speed and angle with respect to the horizontal of the velocity vector at impact

2a) throwing the stone with more force (1.5 times harder) multiplies the time it takes to reach the ground by?

2b) The horizontal component of the velocity is multiplied by what factor?

2c)how many times further does the stone land away from the building

Homework Equations



d=1/2gt^2
Xf=Xi+volt

The Attempt at a Solution



a) I figured out that since it take 3.03 seconds to drop 45m then it must also take 3.03 for it to land when being thrown horizontally. solved using d=1/2gt^2

b) knowing time: 64=volt => 64/3.03=Vo ----> Vo = 21.13m/s

everything else I can't seem to figure out
 
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For part (c), can you determine the vertical component of the velocity of the stone at impact? How about the horizontal component?
 
gneill said:
For part (c), can you determine the vertical component of the velocity of the stone at impact? How about the horizontal component?

The total velocity would be sqrt(ViSinθ^2 + ViCosθ^2) so would it be ViSinθ = -(sqrt(ViCosθ^2))??
 
xtrubambinoxpr said:
The total velocity would be sqrt(ViSinθ^2 + ViCosθ^2) so would it be ViSinθ = -(sqrt(ViCosθ^2))??

Well, you don't know the angle of impact (yet), so that wouldn't be helpful. And the magnitude of the final velocity won't be equal to the initial velocity Vi, since the stone will be gaining energy as it falls from its initial elevation to the ground below.

So, let me offer another viewpoint. You know that the horizontal and vertical components of projectile motion are independent (at least until the flying object hits something!). Consider each component separately. Start with the vertical component. How fast will the stone be traveling vertically having fallen 45m from rest?
 
gneill said:
Well, you don't know the angle of impact (yet), so that wouldn't be helpful. And the magnitude of the final velocity won't be equal to the initial velocity Vi, since the stone will be gaining energy as it falls from its initial elevation to the ground below.

So, let me offer another viewpoint. You know that the horizontal and vertical components of projectile motion are independent (at least until the flying object hits something!). Consider each component separately. Start with the vertical component. How fast will the stone be traveling vertically having fallen 45m from rest?

it would be falling with V = -gt. d=1/2gt^2 would get me 3.03 seconds to fall from rest. so -gt = -29.71m/s that is the vertical component itself? considering the horizontal component always stays the same because no force acts on it to provide any accel.
 
xtrubambinoxpr said:
it would be falling with V = -gt. d=1/2gt^2 would get me 3.03 seconds to fall from rest. so -gt = -29.71m/s that is the vertical component itself? considering the horizontal component always stays the same because no force acts on it to provide any accel.

Right. So now you know both the vertical and horizontal components of the velocity at impact (assuming you can calculate the horizontal component given your distance and time).
What can you determine from that?
 
gneill said:
Right. So now you know both the vertical and horizontal components of the velocity at impact...
What can you determine from that?

that final velocity of impact is sqrt(29.71^2 + 21.12^2) will give me the velocity right before it hits the ground.
^^ I think that is wrong

the only other thing I can think of is that the components can be further broken down to i + j form and then placed into each other to get one final equation?
 
The components are already in "i + j form", since one is vertical and the other horizontal; they are at right angles to each other, and directed along coordinate axes. So essentially they are orthogonal vectors.

So your final velocity magnitude is okay. What about the angle?
 
gneill said:
The components are already in "i + j form", since one is vertical and the other horizontal; they are at right angles to each other, and directed along coordinate axes. So essentially they are orthogonal vectors.

So your final velocity magnitude is okay. What about the angle?

if they are in I + J form already then tan^-1(j/i) gives me the angle measure which is -54.6 degrees?
 
  • #10
xtrubambinoxpr said:
if they are in I + J form already then tan^-1(j/i) gives me the angle measure which is -54.6 degrees?

Sure. Looks good.
 
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  • #11
Got it! took a second look at it and became so obvious why. thank you!
 
  • #12
xtrubambinoxpr said:
Got it! took a second look at it and became so obvious why. thank you!

It's always a good feeling when what looked difficult suddenly becomes obvious :smile:

Cheers!
 

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