Simple kinematics problem I can't figure out!

  • Thread starter MRi04
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  • #1
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Homework Statement


A ball is thrown down with an initial velocity of 2.0 m/s from a height of 1.75 m, determine how long until it hits the ground.

initial velocity = 2 m/s
distance = 1.75 m
acceleration = 9.8 m/s2
time = ?

Homework Equations


d=vit+1/2a(t)2

The Attempt at a Solution


1.75=2t + 1/2(9.8)t2
1.75=2t + 4.9t2
Square root of both sides???
1.32 = 1.4t + 2.21t
1.32 = 3.61t
All divided by 3.61???
t = 0.36
.... I know the solution to be 0.43 seconds, but can't get there!

Any help is greatly appreciated!
 

Answers and Replies

  • #2
RUber
Homework Helper
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Homework Statement


A ball is thrown down with an initial velocity of 2.0 m/s from a height of 1.75 m, determine how long until it hits the ground.

initial velocity = 2 m/s
distance = 1.75 m
acceleration = 9.8 m/s2
time = ?

Homework Equations


d=vit+1/2a(t)2

The Attempt at a Solution


1.75=2t + 1/2(9.8)t2
1.75=2t + 4.9t2
Square root of both sides???
That is where you went wrong...
Solve this as a quadratic equation.
 
  • #3
36
6

Homework Statement


A ball is thrown down with an initial velocity of 2.0 m/s from a height of 1.75 m, determine how long until it hits the ground.

initial velocity = 2 m/s
distance = 1.75 m
acceleration = 9.8 m/s2
time = ?

Homework Equations


d=vit+1/2a(t)2

The Attempt at a Solution


1.75=2t + 1/2(9.8)t2
1.75=2t + 4.9t2
Square root of both sides???
1.32 = 1.4t + 2.21t
1.32 = 3.61t
All divided by 3.61???
t = 0.36
.... I know the solution to be 0.43 seconds, but can't get there!

Any help is greatly appreciated!
In your solution,you took square root of both sides which is incorrect.Instead,solve the quadratic equation formed in t.You will get the time taken.
 
  • #4
6
1
Thanks RUber and Ellispson, really appreciate it, can you give any more detail, I'm not sure how to do that?
 
  • #5
RUber
Homework Helper
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344
If you have an equation of the form : At^2 + Bt + C = 0, then
##t = \frac{ -B \pm \sqrt{B^2 - 4AC}}{2A}##
Only one of the solutions will be a positive time...choose that one.
 
  • #6
6
1
If you have an equation of the form : At^2 + Bt + C = 0, then
##t = \frac{ -B \pm \sqrt{B^2 - 4AC}}{2A}##
Only one of the solutions will be a positive time...choose that one.
but what does C represent in the quadratic equation? Sorry, it has been a long time since I have used the quadratic equation.
 
  • #7
6
1
but what does C represent in the quadratic equation? Sorry, it has been a long time since I have used the quadratic equation.
I mean, what does it represent in terms of the problem I posted?
 
  • #8
RUber
Homework Helper
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344
You have 1.75 = 2t + 4.9t^2 . Can you put that into the form of At^2 + Bt + C = 0? C is the number that is not multiplied by t or t^2...I get -1.75.
 
  • #9
36
6
I mean, what does it represent in terms of the problem I posted?
The constant term.Which means the term not having the variable t in it.
 
  • #10
6
1
You have 1.75 = 2t + 4.9t^2 . Can you put that into the form of At^2 + Bt + C = 0? C is the number that is not multiplied by t or t^2...I get -1.75.
Oh of course, I see. I'll try that now
 
  • #11
6
1
That worked for me! Thanks very much to both of you. Happy physics-ing!
 
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