Respuesta :
The shot was blocked between 0.84 and 0.85 seconds after the shot is launched.
what will be the time in which the blocker will block the ball?
The given equation for ball height
[tex]=6+30t-16t^{2}[/tex]
The equation for the blocker's height will be
[tex]=9+25t-16t^{2}[/tex]
But, the shot is made before two-tenths of a second or 0.2 seconds therefore modified equation for ball height is
[tex]=6+30(t-0.2)-16(t-0.2)^{2}[/tex]
Now for the shot to be blocked, the height of the shot-blocker must be greater than the height of the ball which is shot before 0.2 seconds :
[tex]9+25t-16t^{2} \geq 6+30(t-0.2)-16(t-0.2)^{2}[/tex]
[tex]9+25t-16t^{2} \geq6+30t-6-16(t^{2} -0.4t+0.04)[/tex]
[tex]9+25t-16t^{2} \geq30t-16t^{2} -6.4t+0.64[/tex]
[tex]9+25t\geq36.4t-0.64[/tex]
[tex]9.64\geq11.4t[/tex]
[tex]t \leq \dfrac{9.64}{11.4}=0.846[/tex]
Thus the shot was blocked between 0.84 and 0.85 seconds after the shot is launched.
To know more about the Equation of motions follow
https://brainly.com/question/25951773