Find time in velocity/Gravity situation

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A stone thrown upward with an initial velocity of 20 m/s reaches its highest point in 2 seconds and achieves a maximum height of 20 meters. The calculations involve using the first equation of motion, where the final velocity at the peak is zero, and the acceleration due to gravity is -10 m/s². A common mistake noted in the discussion is confusion over the signs in the equations, particularly when multiplying or dividing by negative numbers. Understanding the direction of vectors is crucial, as upward motion is positive and gravity is negative. Proper handling of these signs is essential for accurate problem-solving in kinematics.
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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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