Sounds like a physics problem
basically we can simlpify it to this:
we toss something straight up with initial velocity 10ft/sec, assuming gravity kicks in, what is max height?
we can use the kinematic equation
![v_{f}^2=v_{0}^2+2ad](https://tex.z-dn.net/?f=v_%7Bf%7D%5E2%3Dv_%7B0%7D%5E2%2B2ad)
![v_f](https://tex.z-dn.net/?f=v_f)
is final velocity
![v_0](https://tex.z-dn.net/?f=v_0)
is initial velocity
a is acceleration due to gravity
d=distance traveled
if we say he tosses it straight up then when it reaches max height,
![v_f=0](https://tex.z-dn.net/?f=%20v_f%3D0)
and
![v_0=10 \frac{ft}{s}](https://tex.z-dn.net/?f=v_0%3D10%20%5Cfrac%7Bft%7D%7Bs%7D)
and we know that
![a=-32.174 \frac{ft}{s^2}](https://tex.z-dn.net/?f=a%3D%3Cspan%3E-32.174%20%5Cfrac%7Bft%7D%7Bs%5E2%7D)
so solving for d
<span>
![v_{f}^2=v_{0}^2+2ad](https://tex.z-dn.net/?f=v_%7Bf%7D%5E2%3Dv_%7B0%7D%5E2%2B2ad)
</span><span>
![v_{f}^2-v_{0}^2=2ad](https://tex.z-dn.net/?f=v_%7Bf%7D%5E2-v_%7B0%7D%5E2%3D2ad)
</span><span>
![\frac{v_{f}^2-v_{0}^2}{2a}=d](https://tex.z-dn.net/?f=%5Cfrac%7Bv_%7Bf%7D%5E2-v_%7B0%7D%5E2%7D%7B2a%7D%3Dd)
plug the numbers in
</span>
![\frac{(0)^2-(10 \frac{ft}{s})^2}{2(-32.174 \frac{ft}{s^2})}=d](https://tex.z-dn.net/?f=%5Cfrac%7B%280%29%5E2-%2810%20%5Cfrac%7Bft%7D%7Bs%7D%29%5E2%7D%7B2%28%3Cspan%3E-32.174%20%5Cfrac%7Bft%7D%7Bs%5E2%7D%29%7D%3Dd)
![1.5540498539193137315845092310561 ft=d](https://tex.z-dn.net/?f=1.5540498539193137315845092310561%20ft%3Dd)
we add that to the initial 6ft
so total of 7.5540498539193137315845092310561 ft max height
about 7.6ft
Answer:
m = 5x – 8
m = 3y,
m = 132
Since m = 3 y then y = 132/3
then y = 44
132 = 5x - 8
5 x = 140
x = 28
Step-by-step explanation:
All the vertices of a triangle are equidistant from the circumcenter
Do you have to create an equation?
Answer: 5z^(-5/3)
Step-by-step explanation:
Given the expression above, rewrite in the form k × z^n
(10 × (z)^1/3) ÷ 2z^2
(10z^1/3) ÷ (2z^2)
10/2(z^(1/3 - 2))
5z^(-5/3)