From the top of a building 48 feet high, a ball is thrown upward with an initial velocity of 48 feet per second.

Question

From the top of a building 48 feet high, a ball is thrown upward with an initial velocity of 48 feet per second.

(a) When does it reaches its maximum height? when t= ____ seconds

(b) What is its maximum height? ____ feet

(c) When does it hit the ground? when t= ____ seconds

(d) With what speed does it hit the ground? _____ feet per second (a negative number)

(e) What is its acceleration at t=2 ? ______ feet per second per second (a negative number)

Answer 1

h(t) = -16t^2 + 48t + 48

v(t) = h'(t) = -32t +48 = 0

t = 48/32 = 3/2 seconds to reach max height

max height = -16(3/2)^2 + 48(3/2) + 48= -36 + 72 +48 = 84 feet high

it hits the ground when 0=-16t^2 +48t + 48

t^2 -3t -3 = 0

t^2-3t +9/4 = 3+9/4

(t-3/2)^2 = 21/4

t-3/2 = +/- sqr21/2

t = (3+sqr21)/2 = about 3.79 seconds to hit the ground

its velocity at about 3.79 seconds = v(3.79) = -32(3.79)+48= -73.32 feet per second when it hits the ground

acceleration at time t=2 is -32 feet per second per second

the acceleration is a constant from time t=0 to time it hits the ground

reflecting the force of gravity