A ball is dropped from the top of the Empire State building to theground below. The height, y, of the ball above the ground (in feet)is given as a function of time, t (in seconds) by.(A) Find the velocity of the ball at time t. What is the sign ofthe velocity? Why is this to be expected?(B) Show that the acceleration of the ball is a constant. What arethe value and sign of this constant?(C) When does the ball hit the ground, and how fast is it going atthat time? Give your answer in feet per second and in miles perhour (1ft/sec=15/22mph).

Answer:A) [tex]v(t)=v_{0}+at[/tex],B) The aceleration is constant due to [tex]a=g=-9.8\frac{m}{s^{2} }[/tex] C) t=8,81s[tex]86,33\frac{m}{s}*\frac{feet}{0.3048m}=283,23\frac{feet}{s}\\86,33\frac{m}{s}*\frac{3600s}{1h}*\frac{mile}{1609,34m}=193,11\frac{mile}{h}[/tex]Step-by-step explanation:Be y as the heigh (y) of the roof of the empire state (381m)A) [tex]v=v_{0}+a.t[/tex]  to [tex]v_{0}=0[/tex] then v=-a.t, so ([tex]a=-9.8\frac{m}{s^{2} }[/tex])  ; acceleration of gravitysince the acceleration is down  the falling speed is negativeB) The aceleration is constant due to [tex]a=g=-9.8\frac{m}{s^{2} }[/tex] C) [tex]y=y_{0}+v_{0}t+\frac{1}{2}at^{2}[/tex][tex]0=381+0.t+\frac{1}{2}(-9.8)t^{2}[/tex][tex]-9.8t^{2}=-762\\t^{2}=77,75\\t=\sqrt{77,75} =8,81s[/tex][tex]86,33\frac{m}{s}*\frac{feet}{0.3048m}=283,23\frac{feet}{s}\\86,33\frac{m}{s}*\frac{3600s}{1h}*\frac{mile}{1609,34m}=193,11\frac{mile}{h}[/tex]