(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A stone is thrown upward at an angle of 51° above the horizontal. Its maximum height during the trajectory is 26 m. What was the stone's initial speed?

m/s

2. Relevant equations

a=v^{2}/r

v^{2}=v_{o}^{2}+2ax

C=2[tex]\pi[/tex]r

3. The attempt at a solution

I figured the acceleration is 9.81*sin51 since it just gravity when thrown upward so now I have to account for gravity.

so a=7.62

I plugged this into a=v^{2}/r in order to find velocity:

7.62=v^{2}/26

so v=14.079

The distance I got by using the equation C=2[tex]\pi[/tex]r

I got that C=163.36 and the distance is 1/4th of the circumference since the ground and highest point are perpendicular. I got 40.84 for the distance

I then plugged my answers for a, v and distance into v^{2}=v_{o}^{2}+2ax to find initial velocity

I got that v=20.596 but this was incorrect. Does anyone know what I did wrong?

Thank you!

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Free Fall stone physics

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