6.6 miles I think hope this is the right answer.
Answer:
![v = 14.86 m/s](https://tex.z-dn.net/?f=v%20%3D%2014.86%20m%2Fs)
Explanation:
As we know that the force equation at the top is given as
![\frac{mv^2}{R} = ma](https://tex.z-dn.net/?f=%5Cfrac%7Bmv%5E2%7D%7BR%7D%20%3D%20ma)
now we know that
![a_c = 1.5 g](https://tex.z-dn.net/?f=a_c%20%3D%201.5%20g)
so we have
![\frac{v^2}{R} = 1.5 g](https://tex.z-dn.net/?f=%5Cfrac%7Bv%5E2%7D%7BR%7D%20%3D%201.5%20g)
![v = \sqrt{1.5 Rg}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B1.5%20Rg%7D)
so we will have
![v = \sqrt{1.5(15)(9.81)}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B1.5%2815%29%289.81%29%7D)
![v = 14.86 m/s](https://tex.z-dn.net/?f=v%20%3D%2014.86%20m%2Fs)
Explanation:
Step one:
given dara
mass of ball m=0.kg
initial velocity u= 10 m/s
final velocity v=10m/s
Required
momentum before and after impact
the expression for momentum P
P=mv
before impact
P=mv
P1=0.5*10
P1=5(kg⋅m/s)
after impact
P=mu
P2=0.5*10
P2=5(kg⋅m/s)
change in momentum is =P1-P2= 5-5=0
yes, momentum is conserved.