Find time in velocity/Gravity situation

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Homework Help Overview

The discussion revolves around a kinematics problem involving a stone thrown upward with an initial velocity of 20 m/s. Participants are tasked with calculating the time taken for the stone to reach its highest point and the maximum height it achieves.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to determine time and height, with some expressing confusion over the algebraic manipulation of the equations, particularly regarding the signs involved.

Discussion Status

Some participants have provided calculations and attempted to clarify the algebraic steps, while others have raised questions about the handling of negative signs in the equations. There is an ongoing exploration of the implications of these signs on the results.

Contextual Notes

Participants are reminded of forum rules to include relevant equations and attempts at solutions in their posts. There is an acknowledgment of the common difficulties associated with understanding the directionality of vectors in kinematic equations.

nasadall
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Homework Statement


d) A stone is thrown upward with a velocity of 20m/s. Calculate:

i) How long does it take for the stone to reach its highest point;


ii) the maximum height it will achieve?


Please help
 
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The initial velocity, u = 20m/s
The final velocity will be zero at the top most point as the stone will stop at its maximum height
thus, final velocity, v = 0m/s

the acceleration,gravity(negative as it is pulling the stone down) = 10m/s^2 (use 9.8 for accurate results)

by the first equation of motion, v = u + at
=> 0 = 20 +(-10)t
=>-20=-10 t
=>t = 2

by the third equation of motion,
v^2 + u^2 = 2as
=> 0^2 + 20^2 = 2 X -10 X s
on solving,
s = 20 m

thus, maximum height is 20 m and the time taken is 2 seconds

( 10^2 means 10 squared)
 
AlchemistK said:
The initial velocity, u = 20m/s
The final velocity will be zero at the top most point as the stone will stop at its maximum height
thus, final velocity, v = 0m/s

the acceleration,gravity(negative as it is pulling the stone down) = 10m/s^2 (use 9.8 for accurate results)

by the first equation of motion, v = u + at
=> 0 = 20 +(-10)t
=>-20=-10 t
=>t = 2

by the third equation of motion,
v^2 + u^2 = 2as
=> 0^2 + 20^2 = 2 X -10 X s
on solving,
s = 20 m

thus, maximum height is 20 m and the time taken is 2 seconds

( 10^2 means 10 squared)

Thank you, but on the first question, nearly at the end, don't the -10 when it crosses over to the left stays as -20/10=t ?? 'couse like that iwould get an answer of -2s, what am i doing wrong
 
Your getting hung up on the algebra and plus/minus sign when you multiply or divide.

If - 20 = - 10t, then multiply both sides by (-1)

(-20)(-1) = (-10t)(-1)

+ 20 = +10t

t = +2 seconds
 
PhanthomJay said:
Your getting hung up on the algebra and plus/minus sign when you multiply or divide.

If - 20 = - 10t, then multiply both sides by (-1)

(-20)(-1) = (-10t)(-1)

+ 20 = +10t

t = +2 seconds

Man, what a minor detail, do you believe i got stuck in most of my equations because of that.

I need to go back to basics. lololol

Thank you very much
 
You are not the first to get stung by the minus sign. :smile:

In the future, please post you relevant equations and attempt at a solution. These are forum rules, thanks.

You should note in the kinematic equations of motion , that

vf = vo + at, for example , and as another example

vf2 = vo2 + 2as

Note that if you choose up as the positive direction, then any vector that acts up (such as the displacement and initial velocity in this problem) gets a positive sign , and any vector that acts down (like gravity in this problem) gets a negative sign. It's one thing to mess up the minus sign in algebra, and another to get it messed up in the understanding of the actual physics of the problem.

Unfortunately, the minus sign will bite you again and again and again...:frown:
 

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