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