McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
Study Guide and Review
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Exercise 73 Page 84

Let d=64 and solve for t using the equation provided.

2 sec

Practice makes perfect
The water is 64 ft below, so the boulder has to travel a distance of 64 ft. Let's substitute that for d in our equation, then solve for t.
d=16t^2
64=16t^2
â–Ľ
Solve for t
4 = t^2
±2 = t
t = ±2
Since negative time does not make sense in this scenario, we can ignore the negative solution and say t=2. It will take the boulder 2 sec to hit the water.