Answer:
Following are the solution to the given points:
Explanation:
For point a:
Find the schematic of the empty body and in attachment. Upon on ball during the pitch only two forces act:
The strength of the pitcher F is applied that operates horizontally. Its gravity force acting on an object is termed weight, which value is where m denotes mass, and g the acceleration of gravity.
For point b:
[tex]160.2\ N[/tex]
First, they must find that ball's acceleration. You can use the SUVAT equation to achieve that
where
[tex]v = 47\ \frac{m}{s} \\\\u = 0 \\\\a =?\\\\d = 1.0 \ m \\\\[/tex]
Solving for a,
[tex]a=\frac{v^2-u^2}{2d}=\frac{47^2-0}{2(1.0)}=1104.5 \ \frac{m}{s^2}[/tex]
Calculating the mass:
[tex]m = 145 g = 0.145 kg[/tex]
Calculating the force:
[tex]F=ma=0.145 \times 1104.5= 160.2 \ N[/tex]
For point c:
0.195 times the pitcher's weight
[tex]m = 84 \ kg \\\\g = 9.8\ \frac{m}{s^2}\\\\[/tex]
Solving for W:
[tex]W=84 \times 9.8= 823.2 \ N[/tex]
Now the force of Part B could be defined as the fraction of the mass of the pitcher:
[tex]\frac{F}{W}=\frac{160.2}{823.3}=0.195[/tex]
