An amusement park

An amusement park has a ride with a free fall of 128 feet. The formula t=Sqrt(2d/g) gives the time t in seconds it takes the ride to fall a distance of d feet.  The formula v= sqrt(2gd) gives the velocity v in feet per second after the ride has fallen d feet. The letter g represents the gravitational constant.  

A. Rewrite each formula so that the variable d is isolated. Then simplify each formula using the fact the g~32ft/s^2. 

B. Find the time it takes the ride to fall halfway and its velocity at that time.  Then find the time and velocity for the full drop. 

C. What is the ratio of the time it takes for the whole drop to the time it takes for the first half? What is the ratio of the velocity after the second half of the drop to the velocity after the first half? What do you notice

(A)  t=Sqrt(2d/g)        square both sides                       v =     sqrt(2gd)

t^2  = 2d/ g    multiply both sides by  g                         v^2  = 2gd       divide both sides by 2g

t^2/ g = d  → t^2 / [ -32ft/s^2]  = d                                v^2 /  [2g] = d    →   v^2/ [2 * -32ft/s^2] = d

(B)    t = sqrt [ 2(-64)/(-32) ]  = 2 sec to fall halfwav

        v = sqrt(2 * -32 * -64)  =  64 ft/s

(C) t = sqrt[(2(-128)/(-32) ]  = 2sqrt(2)  sec to fall completely........the ratio  of this time to the halfway drop time = sqrt(2)  times as long

v = sqrt [ 2(-128)* (-32) ]  = 64sqrt(2) ft/s = terminal velocity......this is sqrt(2) times as great as the halfway velocity........both the time and the velocity ratios are exactly the same

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