it's 1727 +822 just kidding
The element would have 52 protons because atomic number gives protons.
Answer:
hmax=81ft
Explanation:
Maximum height of the object is the highest vertical position along its trajectory.
The vertical velocity is equal to 0 (Vy = 0)
![0=V_{y}-g*t=v_{0}*sin(\alpha)-g*th\\](https://tex.z-dn.net/?f=0%3DV_%7By%7D-g%2At%3Dv_%7B0%7D%2Asin%28%5Calpha%29-g%2Ath%5C%5C)
we isolate th (needed to reach the maximum height hmax)
![th = \frac{v_{0}*sin(\alpha)}{g}](https://tex.z-dn.net/?f=th%20%3D%20%5Cfrac%7Bv_%7B0%7D%2Asin%28%5Calpha%29%7D%7Bg%7D)
The formula describing vertical distance is:
![y = Vy * t-g* t^{2} / 2](https://tex.z-dn.net/?f=y%20%3D%20Vy%20%2A%20t-g%2A%20t%5E%7B2%7D%20%2F%202)
So, given y = hmax and t = th, we can join those two equations together:
![hmax = Vy* th-g*th^{2}/2](https://tex.z-dn.net/?f=hmax%20%3D%20Vy%2A%20th-g%2Ath%5E%7B2%7D%2F2)
![hmax =Vo^{2}*sin(\alpha )^{2}/(2*g)](https://tex.z-dn.net/?f=hmax%20%3DVo%5E%7B2%7D%2Asin%28%5Calpha%20%29%5E%7B2%7D%2F%282%2Ag%29)
if we launch a projectile from some initial height h all you need to do is add this initial elevation
![hmax =h+Vo^{2}*sin(\alpha)^{2}/(2*g)](https://tex.z-dn.net/?f=hmax%20%3Dh%2BVo%5E%7B2%7D%2Asin%28%5Calpha%29%5E%7B2%7D%2F%282%2Ag%29)
![hmax =3+100^{2}*sin(45)^{2}/(2 * 32)=81 ft](https://tex.z-dn.net/?f=hmax%20%3D3%2B100%5E%7B2%7D%2Asin%2845%29%5E%7B2%7D%2F%282%20%2A%2032%29%3D81%20ft)