Answer:
Step-by-step explanation:
Given
initial Velocity(u)=44 ft/s
ball initial height =4 ft
as launch angle is not given so considering ball is hit vertically upward
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
where u =initial velocity
v=final velocity
a=acceleration
s=distance traveled
![0-44^2=2\times (-32.2)\cdot s](https://tex.z-dn.net/?f=0-44%5E2%3D2%5Ctimes%20%28-32.2%29%5Ccdot%20s)
![s=\frac{44^2}{2\times 32.2}=30.06 ft](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B44%5E2%7D%7B2%5Ctimes%2032.2%7D%3D30.06%20ft)
time taken to reach 30.06 ft
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
![0=44-32.2\times t](https://tex.z-dn.net/?f=0%3D44-32.2%5Ctimes%20t)
![t=1.36 s](https://tex.z-dn.net/?f=t%3D1.36%20s)
total distance to reach ground=4+30.06 ft=34.06 ft
![s=ut+\frac{at^2}{2}](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7Bat%5E2%7D%7B2%7D)
![34.06=\frac{32.2\times t^2}{2}](https://tex.z-dn.net/?f=34.06%3D%5Cfrac%7B32.2%5Ctimes%20t%5E2%7D%7B2%7D)
![t^2=2.115](https://tex.z-dn.net/?f=t%5E2%3D2.115)
t=1.45 s
I think it is 53.54 because if you divide 40 by 3 and then multiply that by 4
3217 + 13.1 + 1.3 can also be written as
3217.0
+ 13.1
1.3
----------
<span>3231.4
</span>
Your final answer should be <span>3231.4 or option 4. Hope this helps!</span>