If you use reverse propulsion then it will be able to fly with more ease in the vacuum because of the reverse propulsion
Answer:

Step-by-step explanation:
The motion equations that describe the ball are, respectively:
![x = \left[\left(152\,\frac{ft}{s} \right)\cdot \cos 52^{\circ} \right] \cdot t](https://tex.z-dn.net/?f=x%20%3D%20%5Cleft%5B%5Cleft%28152%5C%2C%5Cfrac%7Bft%7D%7Bs%7D%20%5Cright%29%5Ccdot%20%5Ccos%2052%5E%7B%5Ccirc%7D%20%5Cright%5D%20%5Ccdot%20t)
![y = 4.5\,ft + \left[\left(152\,\frac{ft}{s} \right)\cdot \sin 52^{\circ} \right] \cdot t - \frac{1}{2}\cdot \left(32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}](https://tex.z-dn.net/?f=y%20%3D%204.5%5C%2Cft%20%2B%20%5Cleft%5B%5Cleft%28152%5C%2C%5Cfrac%7Bft%7D%7Bs%7D%20%5Cright%29%5Ccdot%20%5Csin%2052%5E%7B%5Ccirc%7D%20%5Cright%5D%20%5Ccdot%20t%20-%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%2832.174%5C%2C%5Cfrac%7Bft%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%20%5Ccdot%20t%5E%7B2%7D)
The time required for the ball to hit the ground is computed from the second equation. That is to say:
![4.5\,ft + \left[\left(152\,\frac{ft}{s} \right)\cdot \sin 52^{\circ} \right] \cdot t - \frac{1}{2}\cdot \left(32.174\,\frac{m}{s^{2}} \right) \cdot t^{2} = 0](https://tex.z-dn.net/?f=4.5%5C%2Cft%20%2B%20%5Cleft%5B%5Cleft%28152%5C%2C%5Cfrac%7Bft%7D%7Bs%7D%20%5Cright%29%5Ccdot%20%5Csin%2052%5E%7B%5Ccirc%7D%20%5Cright%5D%20%5Ccdot%20t%20-%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%2832.174%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%20%5Ccdot%20t%5E%7B2%7D%20%3D%200)
Given that formula is a second-order polynomial, the roots of the equation are described below:
and 
Just the first root offers a realistic solution. Then,
.
Answer:
10
Step-by-step explanation:
A rational number is 9.
When you multiply 0.9x10 you get the sum of 9.
9 is a rational number because it is a perfect square and an integer.
I hope this helps you understand :)
Answer:
2614.33333333
Step-by-step explanation:
well, this is just a matter of simple unit conversion, so let's recall that one revolution on a circle is just one-go-around, radians wise that'll be 2π, and we also know that 1 minute has 60 seconds, let's use those values for our product.
![\cfrac{300~~\begin{matrix} r \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{~~\begin{matrix} min \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{2\pi ~rad}{~~\begin{matrix} r \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{~~\begin{matrix} min \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{60secs}\implies \cfrac{(300)(2\pi )rad}{60secs}\implies 10\pi ~\frac{rad}{secs}\approx 31.42~\frac{rad}{secs}](https://tex.z-dn.net/?f=%5Ccfrac%7B300~~%5Cbegin%7Bmatrix%7D%20r%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7B~~%5Cbegin%7Bmatrix%7D%20min%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%5Ccdot%20%5Ccfrac%7B2%5Cpi%20~rad%7D%7B~~%5Cbegin%7Bmatrix%7D%20r%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%5Ccdot%20%5Ccfrac%7B~~%5Cbegin%7Bmatrix%7D%20min%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7B60secs%7D%5Cimplies%20%5Ccfrac%7B%28300%29%282%5Cpi%20%29rad%7D%7B60secs%7D%5Cimplies%2010%5Cpi%20~%5Cfrac%7Brad%7D%7Bsecs%7D%5Capprox%2031.42~%5Cfrac%7Brad%7D%7Bsecs%7D)