Verified answer

d=vt+0.5at^2

The v is actually the initial velocity, and it's 0 so it cancels out the term.

d= 0.5at^2

Solving for t,

t=sqrt(d/0.5a)

64 feet is 19.5072 metres so and the acceleration on earth is -9.81

t=sqrt(-19.5072/0.5*-9.81)

t= 1.994242477 sec

t= 2 seconds

x^2+4=20

x^2-16=0

This is a difference of squares.

(x+4)(x-4)=0

x=4 or x=-4

You know that the height of the ball is 64 feet.

You know that the acceleration of the ball will be 32.2 ft/s²

You know that the initial velocity is zero because the ball was dropped from "rest"

Therefore, this formula can be used:

d = ut + (0.5)(a)(t^2)

where u = initial velocity = 0 ft/s

t = time (unknown)

d = 64 feet

a = 32.2ft/s^2

Plug in the numbers and this will happen

64 = 0 + (0.5)(32.2)(t^2)

t^2 = 64/16.1 = 3.975

t = 1.99 seconds

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x^2 + 4 = 20

x^2 = 16

x = +4 and -4