How High Does a Ball Go If Thrown Upwards at 40m/s?

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To determine the height reached by a ball thrown upwards at 40 m/s, the relevant equation is V^2 = u^2 + 2as, where u is the initial velocity and a is the acceleration due to gravity (g = -9.8 m/s²). The ball's velocity decreases to zero at its peak height, allowing for the calculation of displacement (s) using the equation rearranged as 0 = 40² + 2*(-g)*s. The time taken for the ball to reach this height can also be calculated using the formula t = V/g, where V is the initial velocity. It's important to note that the initial velocity (u) is 40 m/s, not zero, and that g is expressed in m/s². The discussion emphasizes correctly applying the equations of motion to find the maximum height.
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Homework Statement



Height reached by a ball if it is thrown vertically upwards with an initial velocity of 40m/s

Homework Equations



Using V^2=u^2+2as

The Attempt at a Solution

I have been told to ignore air resistance and let g =9.8 m/s. Any help anyone?
 
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Well you have the correct equation. Just plug in all the variables and solve for the displacement s.
 
Consider that the balls's velocity of 40 m/s is being sapped away by gravity at a rate of 9.8 m/s/s. How much time (t) will it take for v to be reduced to zero?

What will the ball's average velocity be over time period t? (Now you can solve for h.)

(Also: you probably just made a typo but note that g is an acceleration and as such takes the unit "m/s/s", normally written "m/s2". "m/s" is a unit of velocity, not acceleration.).
 
Thank you! Yes it was a typo! I am right that even if this question states initial velocity is 40m/s the actual value of u in this question is zero? Also my transposition of this equation works out to be V^2-u^2 divided by 2a=s?
 
Yes the final velocity is 0 for that equation.

So for your equation:

0 = 402 +2*(-g)*x

(if +X is up, then g carries a - sign.)

If time is all you want then you have an easier path with

|V| = |g|*t

t = V/g
 
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