Aball is thrown vertically upward from the top of a building 160 feet tall with an initial velocity of 48 feet per second. the distance d (in feet) of the ball from the ground after t seconds is d(t) = 160 + 48t – 16t2. a. after how many seconds does the ball strike the ground? write your answer in a complete sentence with proper grammar and correct spelling. b. when will the ball reach its maximum height?

Answer

given,

height of the building = 160 ft

initial velocity = 48 ft/s

the relation of distance is given

d = -16 t²  + 48 t + 160

time at which the ball hit the ground will be when d = 0

-16 t²  + 48 t + 160 = 0

solving quadratic equation

t = 5 s(neglecting the negative solution)

b) ball will reach max height when velocity is equal to zero

    v = 0

     t = 1.5 s

at 1.5 sec the ball will reach at maximum height

f is the midpoint of

i cant answer this, theres no table. sorry man

step-by-step explanation: