Projectile motion question and minimum initial speed

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Homework Help Overview

The discussion revolves around a projectile motion problem involving a historical anecdote about George Washington throwing a silver dollar across a 75m wide river. Participants are tasked with determining the minimum initial speed required for the coin to cross the river and the duration of its flight, given the acceleration due to gravity as 9.8 m/s².

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants express uncertainty about how to approach the problem with two variables and consider the optimal angle for the throw. There are discussions about the relationship between horizontal distance and height, as well as attempts to derive equations for flight time and initial speed.

Discussion Status

Some participants have provided guidance on the relationships between the variables involved, while others are exploring different interpretations of the problem. There is an ongoing exchange about how to calculate maximum height and the implications of assuming a 45-degree launch angle.

Contextual Notes

Participants are grappling with the definitions of variables and the setup of the problem, including the maximum height in relation to the river's width. There is also mention of previous discussions that may influence current reasoning.

justsomebody
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Homework Statement



Assume acceleration due to gravity is 9.8ms^-2

It has been said that in his youth George Washington threw a silver dollar
across a river. Assuming that the river was 75m wide,

(a) what minimum initial speed was necessary to get the coin across the river
(b) how long was the coin in flight?
Ans (a) 27.1 m/s (b) 3.91 s

2. The attempt at a solution

i dun even know where to start with 2 variables =(
only tried assuming his height =o
 
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justsomebody said:

Homework Statement



Assume acceleration due to gravity is 9.8ms^-2

It has been said that in his youth George Washington threw a silver dollar
across a river. Assuming that the river was 75m wide,

(a) what minimum initial speed was necessary to get the coin across the river
(b) how long was the coin in flight?
Ans (a) 27.1 m/s (b) 3.91 s

2. The attempt at a solution

i dun even know where to start with 2 variables =(
only tried assuming his height =o


This is a twist on your question from yesterday. To start this one, ask yourself what the optimal angle would be to throw a coin (assuming no air resistance). Then look at the formulas that you know to see if you can determine the flight time.
 
Borg said:
This is a twist on your question from yesterday. To start this one, ask yourself what the optimal angle would be to throw a coin (assuming no air resistance). Then look at the formulas that you know to see if you can determine the flight time.

=O u remembered
thx for the help again =D

Assuming angle to be 45 degrees,
maximum height i tink is => tan45 = h / 75
h = 75m
s = ut + 1/2at^2
t = sqrt(75/4.9)
= 3.912303982
= 3.91s which answered part b

but is there anyway to get part a 1st before part b =O
mayb there is a way now with the angle =o

is this wrong?
i use v^2 = u^2 + 2as for the vertical velocity
v = 0, when it hit the water,
u^2 = 2(9.8)(75)
u = sqrt(ANS)
= 38.34057903m/s

then i tought that if 45 degrees, Vx = Vy, initial velocity u = sqrt(2Vy^2)
u = sqrt(2*38^2)
u = 54.22176685 which is twice of the answer =o
 
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justsomebody said:
=O u remembered
thx for the help again =D

Assuming angle to be 45 degrees,
maximum height i tink is => tan45 = h / 75
h = 75m
s = ut + 1/2at^2
t = sqrt(75/4.9)
= 3.912303982
= 3.91s which answered part b

but is there anyway to get part a 1st before part b =O
mayb there is a way now with the angle =o

Hi again. You got the right answer but, you mislabeled one item. The maximum height is not 75m - that is the distance across the river. The equation for the X-direction is:

s = ut where u is the velocity in the x direction

In the s = ut + 1/2at^2 equation, u is the vertical direction. But, since both velocities start with the same value, you know that ut = 75 from the first equation.

As far as solving it to get the time last, I think they do that to throw you off.
 
Borg said:
Hi again. You got the right answer but, you mislabeled one item. The maximum height is not 75m - that is the distance across the river. The equation for the X-direction is:

s = ut where u is the velocity in the x direction

In the s = ut + 1/2at^2 equation, u is the vertical direction. But, since both velocities start with the same value, you know that ut = 75 from the first equation.

As far as solving it to get the time last, I think they do that to throw you off.

=o ic

but is it possible to assume the maximum height to be at half of the river's width? =o
 
justsomebody said:
=o ic

but is it possible to assume the maximum height to be at half of the river's width? =o

Yes. Can you tell me what it would be?
 
i guess 37.5m =o both the height and half-width
 
justsomebody said:
i guess 37.5m =o both the height and half-width

Don't guess. :smile: What is the vertical velocity at it's highest point?
 
Borg said:
Yes. Can you tell me what it would be?

justsomebody said:
i guess 37.5m =o both the height and half-width

Borg said:
Don't guess. :smile: What is the vertical velocity at it's highest point?

I think we're confusing each other. I'm asking what the maximum height is and not where the max occurs.
 
  • #10
Borg said:
Don't guess. :smile: What is the vertical velocity at it's highest point?

lol now I am stuck again is it zero?

Borg said:
I think we're confusing each other. I'm asking what the maximum height is and not where the max occurs.

i can't do it =o i only used 37.5 * tan45 lol = 37.5 oo
how do u find maximum height lol
 
  • #11
justsomebody said:
lol now I am stuck again is it zero?

What goes up, has to come down. Remember your first two equations. One applies to the horizontal direction and the other to the vertical. [STRIKE]What is s when u = 0?[/STRIKE]
 
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  • #12
Borg said:
What goes up, has to come down. Remember your first two equations. One applies to the horizontal direction and the other to the vertical. What is s when u = 0?

first equation => s = ut (horizontal)
second equation => s = ut + 1/2at^2 ( vertical )
s = 1/2at^2? =o
but i still need to find t 1st =o

man I am confused now i dun even know how to find t =o
 
  • #13
justsomebody said:
first equation => s = ut (horizontal)
second equation => s = ut + 1/2at^2 ( vertical )
s = 1/2at^2? =o
but i still need to find t 1st =o

I was afraid that I would confuse you with that last question. Sorry, let's take a step back. Think about your problem from yesterday. Yesterday, you knew the height and wanted to solve for time. Now you have the time (remember - halfway across) and you want to know the height. Remember that yesterday's initial vertical velocity was zero?
 
  • #14
Borg said:
I was afraid that I would confuse you with that last question. Sorry, let's take a step back. Think about your problem from yesterday. Yesterday, you knew the height and wanted to solve for time. Now you have the time (remember - halfway across) and you want to know the height. Remember that yesterday's initial vertical velocity was zero?

lol I am sorry for taking up your time =o

but with the equation s = 1/2at^2, as u = 0.
s = 1/2(9.8)(3.91)^2 which = 74.91169 ~ 75m o_O

im still guessing the t should be divided by 2
den max h = 75/2 = 37.5 =O
=o nvm u mentioned halfway across =D

is it correct =D
 
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  • #15
justsomebody said:
lol its not your fault i put part of the blame on my lecturer for not teaching projectile motion =o

but with the equation s = 1/2at^2, as u = 0.
s = 1/2(9.8)(3.91)^2 which = 74.91169 ~ 75m o_O

im still guessing the t should be divided by 2
den max h = 75/2 = 37.5 =O
=o nvm u mentioned halfway across =D

I'm new at this as well but, hang in there and we'll get finished.

Take another look at s = ut + 1/2at^2. What are these things? s is the height above the ground. You solved the previous equation because you knew that when the coin returned to earth, s = 0. Now you want to solve for s at the halfway point (t = 3.91s/2). From what you have solved already, you know that the initial velocity is 27.1. Therefore, you have the following:

s = 27.1t + (1/2)(-9.8)t^2

Plug in t for the halfway point.
 
  • #16
its ok ill try to clear my mind and do it at a later time thank you so much for all your help =D
 
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  • #17
justsomebody said:
its ok ill try to clear my mind and do it at a later time thank you so much for all your help =D

I think that I really confused you with the zero velocity comment. The maximum height is 34.25 as you've calculated.
 
  • #18
o ok now i just need to redo part a and b can't find height without these 2 anyway lol

im off to bed cya =D and i really appreciate you taking time out to help me =D sometimes too much just isn't enough coz i really need time to digest all this
 
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