Use the position function s(t) = -16t + v_0t + s_0 for free falling objects. A ball is thrown straight down from the top of a 600-foot building with an initial velocity of -30 feet per second. (a) Determine the position and velocity functions for the ball. (b) Determine the average velocity on the interval [1, 3]. (c) Find the instantaneous velocities when t=1 and t=3. (d) Find the time required for the ball to reach ground level. (e) Find the velocity of the ball at impact.

a) v = -30 - 32 t ,  s (t) = 600 - 30 t -16 t² , b) v = -32 ft / s

c) v (1) = -62 ft / s,  v (3) = -126 ft / s , d) t = 7.13 s , e)  v = -258.16 ft / s

Explanation:

a) For this exercise they give us the function of the position of the ball

          s (t) = s (o) + v_o t - 16 t²

notice that you forgot to write the super index

indicate the initial position of the ball

        s (o) = 600 ft

also indicates initial speed

        v_o = - 30 ft / s

let's substitute in the equation

        s (t) = 600 - 30 t -16 t²

to find the speed we use

       v = ds / dt

       v = v_o - 32 t

       v = -30 - 32 t

b) To find the average speed, look for the speed at the beginning and end of the time interval

t = 1 s

     v (1) = -30 -32 1

     v (1) = - 62 ft / s

t = 3 s

     v (3) = -30 -32 3

     v (3) = -126 ft / s

the average speed is

    v = (v (3) -v (1)) / (3-1)

    v = (-126 +62) / 2

    v = -32 ft / s

c) instantaneous speeds, we already calculated them

    v (1) = -62 ft / s

    v (3) = -126 ft / s

d) the time to reach the ground

in this case s = 0

    0 = 600 - 30 t -16 t²

     t² + 1,875 t - 37.5 = 0

we solve the quadratic equation

     t = [-1,875 ±√ (1,875² + 4 37.5)] / 2

     t = [1,875 ± 12.39] / 2

     t₁ = 7.13 s

     t₂ = negative

Since the time must be positive, the correct answer is t = 7.13 s

e) the speed of the ball on reaching the ground

     v = -30 - 32 t

     v = -30 - 32 7.13

      v = -258.16 ft / s

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j