Math finding the distance in time

Question

neglecting air resistance , the distance d that the object falls varies directly as the square of the time t it has failed. An object falls 64 feet in 2 seconds. determine the distance it will fall in 7 seconds

Answer 1

Given :

Acceleration : A (t) = Gravity = 32.174 ft/sec²

Velocity : V (t) = At + V₀

Distance : D (t) = (A/2)*t² + V₀*t + D₀

Assume D₀ = 0 (you are measuring the distance starting from "0")

D (2) = 64 = (32.174/2)*(2)² + V₀*t + 0

64 ~ 64.348 + V₀*t

based on the above, it is safe to assume that they used a Gravity constant slightly different than "standard" or they rounded. Also, it is safe to assume V₀=0

With D₀ = 0 & V₀ = 0 (which are the initial distance & initial velocity prior to any acceleration), the Distance formula becomes :

D = (A/2)*t²

Since we know their 'A' is not exactly gravity, let's figure out what their 'A' is :

D (2) = 64 = (A/2)*(2)²

64 = (A)*(2)

32 = A

:: Check work :: 64 = (32/2)*2²

::::::::::::::::::::::::: 64 = 64 Correct

Now that we know their "A" is not really gravity, but is rather 32 ft/sec², let's plug it into the D formula using '7' as the Time to fall :

D (t) = (A/2)*t²

D (7) = (32/2)*7²

D (7) = 16*49

D (7) = 784 ft