To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to
![v_x = \frac{x}{t}](https://tex.z-dn.net/?f=v_x%20%3D%20%5Cfrac%7Bx%7D%7Bt%7D)
![v_x = \frac{67}{4.5}](https://tex.z-dn.net/?f=v_x%20%3D%20%5Cfrac%7B67%7D%7B4.5%7D)
![v_x = 14.89m/s](https://tex.z-dn.net/?f=v_x%20%3D%2014.89m%2Fs)
The vertical component of velocity is
![-h = v_y t -\frac{1}{2} gt^2](https://tex.z-dn.net/?f=-h%20%3D%20v_y%20t%20-%5Cfrac%7B1%7D%7B2%7D%20gt%5E2)
Here,
h = Height
g = Gravitational acceleration
t = Time
= Vertical component of velocity
![-1.23 = v_y(4.5)-\frac{1}{2} (9.8)(4.5)^2](https://tex.z-dn.net/?f=-1.23%20%3D%20v_y%284.5%29-%5Cfrac%7B1%7D%7B2%7D%20%289.8%29%284.5%29%5E2)
![-1.23= 4.5v_y - 99.225](https://tex.z-dn.net/?f=-1.23%3D%204.5v_y%20-%2099.225)
![v_y = 21.77m/s](https://tex.z-dn.net/?f=v_y%20%3D%2021.77m%2Fs)
The direction of the velocity will be given by the tangent of the components, then
![tan\theta = \frac{v_y}{v_x}](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%20%5Cfrac%7Bv_y%7D%7Bv_x%7D)
![\theta = tan^{-1} (\frac{21.77}{14.89})](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%28%5Cfrac%7B21.77%7D%7B14.89%7D%29)
![\theta = 55.59\°](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2055.59%5C%C2%B0)
The magnitude is given vectorially as,
![|V| = \sqrt{v_x^2+v_y^2}](https://tex.z-dn.net/?f=%7CV%7C%20%3D%20%5Csqrt%7Bv_x%5E2%2Bv_y%5E2%7D)
![|V| = \sqrt{14.89^2 +21.77^2}](https://tex.z-dn.net/?f=%7CV%7C%20%3D%20%5Csqrt%7B14.89%5E2%20%2B21.77%5E2%7D)
![|V| = 26.37m/s](https://tex.z-dn.net/?f=%7CV%7C%20%3D%2026.37m%2Fs)
Therefore the angle is 55.59° and the velocity is 26.37m/s
The time taken by traveler to cover the distance is,
![t=\frac{d}{v}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bd%7D%7Bv%7D)
Substitute the known values,
![\begin{gathered} t=\frac{(11000\text{ ft)}}{(84\text{ mph)(}\frac{1.46667\text{ ft/s}}{1\text{ mph}})_{}} \\ \approx89.3\text{ s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B%2811000%5Ctext%7B%20ft%29%7D%7D%7B%2884%5Ctext%7B%20mph%29%28%7D%5Cfrac%7B1.46667%5Ctext%7B%20ft%2Fs%7D%7D%7B1%5Ctext%7B%20mph%7D%7D%29_%7B%7D%7D%20%5C%5C%20%5Capprox89.3%5Ctext%7B%20s%7D%20%5Cend%7Bgathered%7D)
Therefore, the time taken by traveler to cover the distance is 89.3 s.
Answer:
A 3 feet radius snowball will melt in 54 hours.
Explanation:
As we can assume that the rate of snowball takes to melt is proportional to the surface area, then the rate for a 3 feet radius will be:
T= A(3 ft)/A(1 ft) * 6 hr
A is the area of the snowballs. For a spherical geometry is computing as:
A=4.pi.R^2
Then dividing the areas:
A(3 feet)/A(1 foot) = (4 pi (3 ft)^2)/(4 pi (1 ft)^2) = (36pi ft^2)/(4pi ft^2)= 9
Finally, the rate for the 3 feet radius snowball is:
T= 9 * 6 hr = 54 hr
Neon and helium are both gases that fall under the Noble Gases section in the periodic table of elements.
Hope this helped to answer your question. :D