Answer:
16.3 m/s
Explanation:
We are given that
Mass of golf ball=m=0.045 kg
We have to find the speed right after the hit.
[tex]b_1=(16-1)\times 10^{-3}=15\times 10^{-3} s[/tex]
[tex]b_2=(12-6)\times 10^{-3}=6\times 10^{-3} s[/tex]
Force,F=70 N
Change in momentum=Area of trapezium
Area of trapezium,A=[tex]\frac{1}{2}(b_1+b_2) h[/tex]
Using the formula
[tex]m(v-u)=\frac{1}{2}(15+6)\times 10^{-3}\times 70[/tex]
[tex]0.045(v-u)=\frac{1}{2}(21)\times 10^{-3}\times 70[/tex]
[tex]v-u=\frac{1}{2}\times \frac{1}{0.045}\times 21\times 10^{-3}\times 70[/tex]
[tex] v-u=16.3[/tex]
Initial velocity of golf ball=u=0
[tex]v-0=16.3[/tex]
[tex]v=16.3 m/s[/tex]