Answer:Expression given below
Explanation:
Given mass of spring![\left ( m_1\right )=0.5 kg](https://tex.z-dn.net/?f=%5Cleft%20%28%20m_1%5Cright%20%29%3D0.5%20kg)
Compression in the spring![\left ( x\right )=20 cm](https://tex.z-dn.net/?f=%5Cleft%20%28%20x%5Cright%20%29%3D20%20cm)
Let the spring constant be K
Using Energy conservation
potential energy stored in spring =Kinetic energy of Block![\left ( m_1\right )](https://tex.z-dn.net/?f=%5Cleft%20%28%20m_1%5Cright%20%29)
![\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DKx%5E2%3D%5Cfrac%7B1%7D%7B2%7Dm_1v%5E2)
![v=x\sqrt{\frac{k}{m_1}}](https://tex.z-dn.net/?f=v%3Dx%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm_1%7D%7D)
now conserving momentum
![m_1v=\left ( m_1+m_2\right )v_0](https://tex.z-dn.net/?f=m_1v%3D%5Cleft%20%28%20m_1%2Bm_2%5Cright%20%29v_0)
![v_0=\frac{m_1}{m_1+m_2}v](https://tex.z-dn.net/?f=v_0%3D%5Cfrac%7Bm_1%7D%7Bm_1%2Bm_2%7Dv)
where
is the final velocity
Answer:
![\dfrac{dz}{dt}=0.65\ ft/s](https://tex.z-dn.net/?f=%5Cdfrac%7Bdz%7D%7Bdt%7D%3D0.65%5C%20ft%2Fs)
Explanation:
Given that
x= 150 ft
![\dfrac{dy}{dt}= 7\ ft/s](https://tex.z-dn.net/?f=%5Cdfrac%7Bdy%7D%7Bdt%7D%3D%207%5C%20ft%2Fs)
y= 14 ft
From the diagram
![z^2=x^2+y^2](https://tex.z-dn.net/?f=z%5E2%3Dx%5E2%2By%5E2)
When ,x= 150 ft and y= 14 ft
![z^2=150^2+14^2](https://tex.z-dn.net/?f=z%5E2%3D150%5E2%2B14%5E2)
![z=\sqrt{150^2+15^2}](https://tex.z-dn.net/?f=z%3D%5Csqrt%7B150%5E2%2B15%5E2%7D)
z=150.74 ft
![z^2=x^2+y^2](https://tex.z-dn.net/?f=z%5E2%3Dx%5E2%2By%5E2)
By differentiating with respect to time t
![2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}](https://tex.z-dn.net/?f=2z%5Cdfrac%7Bdz%7D%7Bdt%7D%3D%202x%5Cdfrac%7Bdx%7D%7Bdt%7D%2B2y%5Cdfrac%7Bdy%7D%7Bdt%7D)
![z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}](https://tex.z-dn.net/?f=z%5Cdfrac%7Bdz%7D%7Bdt%7D%3D%20x%5Cdfrac%7Bdx%7D%7Bdt%7D%2By%5Cdfrac%7Bdy%7D%7Bdt%7D)
Here x is constant that is why
![\dfrac{dx}{dt}=0](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdt%7D%3D0)
![z\dfrac{dz}{dt}= y\dfrac{dy}{dt}](https://tex.z-dn.net/?f=z%5Cdfrac%7Bdz%7D%7Bdt%7D%3D%20y%5Cdfrac%7Bdy%7D%7Bdt%7D)
Now by putting the values in the above equation we get
![150.74\times \dfrac{dz}{dt}=14\times 7](https://tex.z-dn.net/?f=150.74%5Ctimes%20%5Cdfrac%7Bdz%7D%7Bdt%7D%3D14%5Ctimes%207)
![\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s](https://tex.z-dn.net/?f=%5Cdfrac%7Bdz%7D%7Bdt%7D%3D%5Cdfrac%7B14%5Ctimes%207%7D%7B150.74%7D%5C%20ft%2Fs)
![\dfrac{dz}{dt}=0.65\ ft/s](https://tex.z-dn.net/?f=%5Cdfrac%7Bdz%7D%7Bdt%7D%3D0.65%5C%20ft%2Fs)
Therefore the distance between balloon and observer increasing with 0.65 ft/s.