A 5.5kg bowling ball initially at rest is dropped from the top of a 12 m building. It hits the ground 1.75s later. Find the net external force on the falling ball?

43.1N

Step-by-step explanation:

this question involves the application of the use of the equations of motion and the newton's law of motion

the parameters are

initial velocity u = 0

distance (s)= 12

mass (m)=5.5kg

t=1.75s

a= acceleration

S= ut + at²

12= (0×1.75) + ( × a × 1.75²)

12= 0 + (1.53125 × a)

12 = 1.53125 × a

a = 12÷1.53125

a= 7.8367m/s²

F=m × a

=5.5 × 7.8367

=43.10N

step-by-step explanation:

circumference divided by 3.141592654 then timesthe radius twice