A stone is thrown off a bridge...

  • #1

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.
 

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
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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?
 
  • #3
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. :oldfrown: 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?
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
How are the equations wrong? my teacher gave us those. :oldfrown: 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.
 
  • #5
PeterO
Homework Helper
2,426
48

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 any one 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.
 

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