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
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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