1-D Motion Problems

  • Thread starter Tchao
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  • #1
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Homework Statement


a. If a ball is launched straight upwards from the ground with an initial velocity of 20.0 m/s, how long does the ball take to reach a height of 15.0 m on the way up?

b. For the same situation, at what time does the ball reach 15.0 m on the way back down?

c. If a ball is dropped from a height of 100.0 m, what is it's height above the ground (in meters) after 3.0 s?

d. If a ball is dropped from a height of 100.0 m, what is it's velocity (in m/s) after 3.00 s? You may treat downward velocity as positive.

Homework Equations


Vx = V0x + axt
x= x0 +v0xt + 1/2axt^2

The Attempt at a Solution


a. 15 = 20t - 4.9t^2
I took the derivative of my equation.
15 = 20 - 9.8t
t = 0.510 s

b. 0 = 20t - 4.9t^2
I assume the height from problem a cancels out with problem b giving 0 and I also took the derivative of equation.
0 = 20 - 9.8t
t= 2.04 s

c. x = 100 - 4.9t^2
Plug in 3.0 s for t.
x = 55.9 m

d. Vx = 9.8 x 3.0 = 29.4 m/s
 

Answers and Replies

  • #2
kuruman
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Hi Tchao and welcome to PF.

I took the derivative of my equation.
Why? Just solve the quadratic and you will get two solutions. What is the meaning of each solution?

On edit: The derivative of 15 with respect to time is not 15.
 
  • #3
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So, I get 3.09 s and 0.990 s when solving using the quadratic equation.
The 0.990 s would be the solution to problem A and the 3.09 s would be the solution to problem.
 
  • #4
kuruman
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So, I get 3.09 s and 0.990 s when solving using the quadratic equation.
The 0.990 s would be the solution to problem A and the 3.09 s would be the solution to problem.
Correct. The two solutions are the two times when the ball is at height 15 m. The last two parts look fine.
 
  • #5
PeroK
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The Attempt at a Solution


a. 15 = 20t - 4.9t^2
I took the derivative of my equation.
15 = 20 - 9.8t
t = 0.510 s
You cannot take the derivative here, because that is an equation for specific values of ##t##.

For example, how long does it take to go ##5m## at ##2m/s##?

You get the equation (1) ##2t = 5## hence ##t = 2.5s##

But, if you differentiate that equation (1), you get the nonsensical ##2 = 0##.

You can only differentiate something that is an equation for all ##t##. For example: ##s = ut + 0.5 at^2## is an equation that holds for all ##t##, where ##u, a## are constants and ##s## is, therefore, a function of ##t##.

Differentiating that equation gives ##v = \frac{ds}{dt} = u + at##, which is then valid for all ##t## and is, as you may recognise, another equation of motion.
 
Last edited:
  • #6
hmmm27
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The only thing left to say is puhlease use BBcode and/or LaTex, both of which are built right in to the editor, are very easy to use, and the links to the guides are right under the input box, on the left.
 
  • #7
kuruman
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The only thing left to say is puhlease use BBcode and/or LaTex, both of which are built right in to the editor, are very easy to use, and the links to the guides are right under the input box, on the left.
Being a first time user, OP may not be familiar with these niceties.
 
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  • #8
hmmm27
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Being a first time user, OP may not be familiar with these niceties.
True, just pointing them out : rather new myself - it's great watching what would normally be indecipherable ascii scribbles turn into textbook-ready formulas.
 
  • #9
hmmm27
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For this example

x= x0 +v0xt + 1/2axt^2
turns into
$$d_t=d_0+v_0t+\frac{1}{2}at^2$$
 
Last edited:

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