Answer:
3.5 seconds
Explanation:
d = 16t^2 + h0
For a falling object, the equation should read:
d(t) = -16t^2 + h(0) [The 16 should have a minus sign, to account for the fact the ball is falling]. h0 we'll assume is actually h(0), or the height at time = 0 (the start). d(t) is the distance from Earth as a function of time, t)
The expression for a free falling object is d = (1/2)gt^2, where g is the acceleration due to gravity and t is time. Since the distance is given in feet, and since (1/2)g = -16, we can assume g is -32 feet/sec^2, the English value for gravity. At time = 0, h is 200 feet (dropped at the start from 200 feet).
Let's assume the term h0 is actually h(t) where h is the height and t is the time, in seconds. h(0) would mean the height at time = 0 seconds, the starting position. h(0) here is 200 feet.
d(t) = -16t^2 + h(0)
d(t) = -16t^2 + 200
We want the value of t such that d(t) = 0 (reaches ground)
d(t) = -16t^2 + 200
0 = -16t^2 + 200 for d(t) = 0
16t^2 = 200
t^2 = 12.5
t = 3.5 seconds