Answer: 2.92 s
Explanation:
Given
Mass of ball is [tex]m=3.9\ kg[/tex]
The initial velocity of the ball is [tex]u=-3.5\ m/s[/tex]
Velocity after the rebound is [tex]v=15.9\ m/s[/tex]
Force during the contact is [tex]F=25.9\ N[/tex]
We know, change in momentum is Impulse
[tex]\Rightarrow F\cdot \Delta t=m(\Delta v)[/tex]
[tex]\Rightarrow 25.9\cdot \Delta t=3.9(15.9-(-3.5))\\\\\Rightarrow \Delta t=\dfrac{3.9\times 19.4}{25.9}=2.92\ s[/tex]
Thus, the force is applied for 2.92 s
