# A stone is thrown off a bridge....

• DracoMalfoy
In summary: I will post the full homework here.In summary, a stone is thrown off of a bridge and reaches a maximum height of 42.87 meters above the water.
DracoMalfoy

## Homework Statement

A stone is thrown off of a bridge that is 40m above the water. The stone lands 48m from the base of the bridge. The stone was thrown with an initial velocity of 15m at 30 degrees above the horizontal.

What is the maximum height the stone reaches above the water? (ans:42.87m)

## Homework Equations

• Vfx=VicosΘ
• Δx=vicosΘ(t)
• Vfy=(VicosΘ)+ay(t)
• Δy=(VisinΘ)(t)+1/2(ay)(t)
• (Vfy)^2=(VisinΘ)^2+2(ay)(Δy)
• Δy=1/2(Vfy+(VisinΘ))(t)
[/B]

## The Attempt at a Solution

ax:13
ay:7.5
Δy:40m
Δx:48m[/B]

I found the x and y-components. Ax:15cos30=13 and Ay:15sin30=7.5.

Then I tried to find the time using the second equation and got 3.69s.

I've been trying to solve for vfy... and I've done everything. I'm not getting that answer. I don't know why I can't understand how to do this. it's frustrating... I even tried to draw it on an axis.

DracoMalfoy said:

## Homework Statement

A stone is thrown off of a bridge that is 40m above the water. The stone lands 48m from the base of the bridge. The stone was thrown with an initial velocity of 15m at 30 degrees above the horizontal.

What is the maximum height the stone reaches above the water? (ans:42.87m)

## Homework Equations

[/B]

• Vfx=VicosΘ
• Δx=vicosΘ(t)
• Vfy=(VicosΘ)+ay(t)
• Δy=(VisinΘ)(t)+1/2(ay)(t)
• (Vfy)^2=(VisinΘ)^2+2(ay)(Δy)
• Δy=1/2(Vfy+(VisinΘ))(t)

## The Attempt at a Solution

ax:13
ay:7.5
Δy:40m
Δx:48m[/B]

I found the x and y-components. Ax:15cos30=13 and Ay:15sin30=7.5.

Then I tried to find the time using the second equation and got 3.69s.

I've been trying to solve for vfy... and I've done everything. I'm not getting that answer. I don't know why I can't understand how to do this. it's frustrating... I even tried to draw it on an axis.

IMHO, some of your equations make no sense; go back to the textbook and look up correct formulas (or go on-line if you do not have a good textbook). Almost all of these types of problems are clones of each other, so once you have done one or two of them you should be able to deal with new ones quickly and efficiently.

What does ##ay## stand for? Why is it equal to 7.5? How and why do you use information about ##x## and/or ##\Delta x## in this problem?

Ray Vickson said:
IMHO, some of your equations make no sense; go back to the textbook and look up correct formulas (or go on-line if you do not have a good textbook). Almost all of these types of problems are clones of each other, so once you have done one or two of them you should be able to deal with new ones quickly and efficiently.

What does ##ay## stand for? Why is it equal to 7.5? How and why do you use information about ##x## and/or ##\Delta x## in this problem?
How are the equations wrong? my teacher gave us those. Δx is the distance that it lands away from the bridge. Δy is the height that its thrown from. I guess that ay and ax are the coordinates?

DracoMalfoy said:
How are the equations wrong? my teacher gave us those. Δx is the distance that it lands away from the bridge. Δy is the height that its thrown from. I guess that ay and ax are the coordinates?

You wrote ##a_x## and ##a_y##, so presumably you know what they mean. I have no prior idea what they stand for.

I could guess that ##a_y## means the y-component of acceleration, since you earlier wrote ##Vf_y = V_i \cos(\theta) + a_y t ,## meaning that the y-component of velocity depends linearly on ##t## with coefficient ##a_y.## Then you go on to say that ##a_y = 7.5,## which could only be true on a much, much smaller planet than Earth, with a much weaker gravitational force. You also say that the initial y-velocity is ##V_i \cos(\theta)##, which cannot possibly be true (if, as you also wrote, the x-component of initial velocity is ##V_i \cos(\theta).## Below that, you claim that ##\Delta y = V_i \sin(\theta) t + \frac{1}{2} a_y t,## and that is false.

I really hope your teacher did not give you the equations exactly as you wrote them.

DracoMalfoy said:

## Homework Statement

A stone is thrown off of a bridge that is 40m above the water. The stone lands 48m from the base of the bridge. The stone was thrown with an initial velocity of 15m at 30 degrees above the horizontal.

What is the maximum height the stone reaches above the water? (ans:42.87m)

## Homework Equations

[/B]

• Vfx=VicosΘ
• Δx=vicosΘ(t)
• Vfy=(VicosΘ)+ay(t)
• Δy=(VisinΘ)(t)+1/2(ay)(t)
• (Vfy)^2=(VisinΘ)^2+2(ay)(Δy)
• Δy=1/2(Vfy+(VisinΘ))(t)

## The Attempt at a Solution

ax:13
ay:7.5
Δy:40m
Δx:48m[/B]

I found the x and y-components. Ax:15cos30=13 and Ay:15sin30=7.5.

Then I tried to find the time using the second equation and got 3.69s.

I've been trying to solve for vfy... and I've done everything. I'm not getting that answer. I don't know why I can't understand how to do this. it's frustrating... I even tried to draw it on an axis.

Not quite sure either of us know what symbols you are using, but when you state
"I found the x and y-components. Ax:15cos30=13 and Ay:15sin30=7.5."
I can say that those two values you have calculated are the horizontal and vertical components of the launch velocity.

Note: if anyone piece of the original information values (48, 40, 15, 30) was deleted, the problem could still be solved. I hope that who ever set the problem made sure that the 4 values are compatible (can all be correct).
If you did not know the 48m answer, the "Ay" value, plus the 40m high bridge would enable you to calculate when the stone will land, then using the "Ax" value you have you could calculate how far away the stone lands - and I just hope that answer is 48 or the whole problem has 4 different answers - one corresponding to the case where the initial values are each "ignored" one at a time.

## 1. How does the height of the bridge affect the distance the stone travels?

The height of the bridge does not directly affect the distance the stone travels. However, a higher bridge may provide a longer trajectory for the stone to travel on, resulting in a longer distance traveled.

## 2. What factors influence the trajectory of the stone?

The trajectory of the stone is influenced by factors such as the initial velocity, the angle at which the stone is thrown, air resistance, and gravity. These factors work together to determine the path the stone will take.

## 3. Why does the stone eventually fall to the ground?

The stone falls to the ground due to the force of gravity acting on it. As the stone travels through the air, the force of gravity pulls it towards the center of the Earth, causing it to eventually fall to the ground.

## 4. How does air resistance affect the motion of the stone?

Air resistance slows down the motion of the stone by exerting a force in the opposite direction of its motion. This force increases as the stone travels at higher speeds, causing it to eventually reach a maximum speed known as terminal velocity.

## 5. Can the distance the stone travels be calculated?

Yes, the distance the stone travels can be calculated using the equation d = v0t + 1/2at2, where d is the distance traveled, v0 is the initial velocity, a is the acceleration due to gravity, and t is the time the stone is in the air. However, this calculation assumes ideal conditions and may vary in real-world scenarios.

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