Aball is thrown directly downward from a cliff 384 feet high with an initial velocity of 48 feet per second. in about how many seconds will it strike the ground? recall that the acceleration due to gravity is 32 feet per second squared.

  about 3.623 seconds

Step-by-step explanation:

The equation for the height can be written as ...

  h(t) = -1/2gt^2 -48t +384 . . . . . where g=32

  h(t) = -16t^2 -48t +384 = -16(t^2 +3t -24)

We want to find t such that h(t) = 0, so ...

  0 = -16(t^2 +3t -24)

  0 = t^2 +3t -24 . . . . . . divide by -16

  0 = t^2 +3t +2.25 -24 -2.25 . . . . . add/subtract a constant to complete the square

  0 = (t +1.5)^2 -26.25 . . . . . rewrite in vertex form

  26.25 = (t +1.5)^2 . . . . . . . add 26.25; next, square root, subtract 1.5

  t = √26.25 -1.5 ≈ 3.623 . . . . seconds

The ball will strike the ground in about 3.623 seconds.

In about 3.62 seconds

Step-by-step explanation:

Convention: downwards positive

s = ut + ½at²

384 = 48t + ½(32)t²

16t² + 48t - 384 = 0

t² + 3t - 24 = 0

Using quadratic formula:

t = [-3 +/- sqrt(3² - 4(1)(-24)]/2(1)

t = [-3 +/- sqrt(105)]/2

t = (-3 + sqrt(105))/2

Or,

3.623474383 seconds

step-by-step explanation:

the first 5 right here the next 5 i will upload as separate question

step-by-step explanation: