Problem
Question 15 If an object is dropped from the top of a 100 foot building, the amount of time $t$ (in seconds) that it takes for the object to be $h$ feet from the ground is given by the formula \[ t=\frac{\sqrt{100-h}}{4} \] How long does it take before the object is 50 feet from the ground? Give an exact answer. It takes seconds for the object to be 50 feet from the ground. How long does it take to reach the ground? (When it is on the ground, $h=0$.) It takes seconds to reach the ground. Question Help: eBook Post to forum Submit Question
Solution
Step 1 ：Substitute \(h=50\) into the given formula to find out how long it takes for the object to be 50 feet from the ground: \(t=\frac{\sqrt{100-50}}{4} = \frac{\sqrt{50}}{4} = \frac{5\sqrt{2}}{4} \approx 1.77 \text{ seconds}\)
Step 2 ：Substitute \(h=0\) into the given formula to find out how long it takes for the object to reach the ground: \(t=\frac{\sqrt{100-0}}{4} = \frac{\sqrt{100}}{4} = \frac{10}{4} = 2.5 \text{ seconds}\)
Step 3 ：So, it takes approximately \(\boxed{1.77}\) seconds for the object to be 50 feet from the ground and \(\boxed{2.5}\) seconds to reach the ground.
