SOLUTION: (setting up the equation) --- an object is dropped from a height (h0) [the zero is subscript.] Then its height after t seconds is given by h=-16t^2+h0, [again, the zero is subscrip

Question 1124050: (setting up the equation) --- an object is dropped from a height (h0) [the zero is subscript.] Then its height after t seconds is given by h=-16t^2+h0, [again, the zero is subscript.] where h is measured in feet.
my question --- if a ball is dropped from 288 feet above the ground, how long does it take to reach ground level?
Thank you :)
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

The starting height is h0 = 288
We want to find the value of t when h = 0, which is when the object hits the ground

h = -16t^2 + h0
h = -16t^2 + 288 <--- replace h0 with the starting height
0 = -16t^2 + 288 <--- replace h with 0
16t^2 = 288
t^2 = 288/16
t^2 = 18
t = sqrt(18) <---- we'll only focus on positive t values
t = 3*sqrt(2) <--- this is the exact solution
t = 4.2426 <--- this is the approximate solution

It takes about 4.2426 seconds for the object to hit the ground. Keep in mind that we are not considering air resistance at all. If the object is light enough (eg: like a feather), then air resistance will play a bigger role here. However, I'm assuming the object dropped is one such that its weight and surface area don't contribute too much that air resistance plays that much of a role.

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