A bowling ball is dropped from the top of a building that has a height of h = 159 m.a) Assuming air resistance is negligible, what is the ball's speed, vf in m/s, when it strikes the ground? b) If drag produces an opposing force of 10 N while the ball's falling, what is the ball's speed, vf in m/s, when it strikes the ground? Assume the ball has a mass of 5 kg.c) A different ball was dropped and the final speed was measured to be vf = 6.5 m/s while the drag force was measured to be 50 Newtons. What is the mass of this ball in kg?

[tex]\begin{gathered} (v_f)^2=(v_i)^2+2gh \\ \\ \text{ we replacing:} \\ \\ (v_f)^2=(0)^2+2(9.8)(159) \\ \\ v_f=\sqrt{3116.4} \\ \\ v_f=55.82\text{ m/s} \end{gathered}[/tex][tex]\begin{gathered} (v_f)^2=(v_i)^2+2gh \\ \\ \text{ we replacing:} \\ \\ (v_f)^2=(0)^2+2(9.8)(159) \\ \\ v_f=\sqrt{3116.4} \\ \\ v_f=55.82\text{ m/s} \end{gathered}[/tex]

b)

To determine the final velocity with the opposing force, we have to determine the net force and then calculate the acceleration that would be used to calculate the velocity, like this:

[tex]\begin{gathered} F_n=mg-10 \\ \\ \text{ we replacing} \\ \\ F_n=5\cdot9.8-10 \\ \\ F_n=49-10=39\text{ N} \\ \\ F=m\cdot a \\ \\ a=\frac{F}{m}=\frac{39}{5} \\ \\ a=7.8\text{ m/s^^b2} \\ \\ \text{ Therefore:} \\ \\ (v_f)^2=(v_i)^2+2ah \\ \\ \text{ we replacing:} \\ \\ (v_f)^2=(0)^2+2(7.8)(159) \\ \\ v_f=\sqrt{2480.4} \\ \\ v_f=49.8\text{ m/s} \end{gathered}[/tex]

c)

We can calculate the mass in the following way, first, we must calculate the acceleration to be able to determine the mass:

[tex]\begin{gathered} (v_{f})^{2}=(v_{i})^{2}+2ah \\ \\ \text{ we replacing:} \\ \\ 6.5^2=0^2+2\cdot a\cdot(159) \\ \\ 318a=42.25 \\ \\ a=\frac{42.25}{318} \\ \\ a=0.133\text{ m/s^^b2} \\ \\ \text{ Now we can calculate the mass using the force balance:} \\ \\ ma=mg-50 \\ \\ a=g-\frac{50}{m} \\ \\ \text{ we replacing:} \\ \\ 0.133=9.8-\frac{50}{m} \\ \\ 1.333-9.8=-\frac{50}{m} \\ \\ -9.667m=-50 \\ \\ m=\frac{-50}{-9.667} \\ \\ m=5.17\text{ kg} \end{gathered}[/tex]