Question

an object is dropped off a building y meters tall. the time t(y) (in seconds) required for the object to hit the ground is given by t(y)=\sqrt{\frac{10y}{49}} find the exact time required for the object to hit the ground if it is dropped off the building in chicago, a height of 70 m. then round the time to the nearest hundredth of a second. part 1 of 2 the exact time required for the object to hit the ground is \square sec.

Explanation:

Step1: Substitute y = 70 into the formula

We are given the formula \( T(y)=\sqrt{\frac{10y}{49}} \), and we need to find the time when \( y = 70 \) meters. So we substitute \( y = 70 \) into the formula:
\( T(70)=\sqrt{\frac{10\times70}{49}} \)

Step2: Simplify the expression inside the square root

First, calculate the numerator: \( 10\times70 = 700 \). Then we have \( \frac{700}{49} \). We can simplify this fraction by dividing both the numerator and the denominator by 7: \( \frac{700\div7}{49\div7}=\frac{100}{7} \). So now the formula becomes \( T(70)=\sqrt{\frac{100}{7}} \)

Step3: Simplify the square root

We know that \( \sqrt{\frac{100}{7}}=\frac{\sqrt{100}}{\sqrt{7}} \), and \( \sqrt{100} = 10 \), so we have \( \frac{10}{\sqrt{7}} \). To rationalize the denominator, we multiply the numerator and the denominator by \( \sqrt{7} \): \( \frac{10\sqrt{7}}{7} \) (this is the exact time).

Step4: Calculate the approximate value (round to the nearest hundredth)

First, calculate \( \sqrt{7}\approx2.6458 \). Then \( 10\sqrt{7}\approx10\times2.6458 = 26.458 \). Then divide by 7: \( \frac{26.458}{7}\approx3.78 \)

Answer:

The exact time is \( \frac{10\sqrt{7}}{7} \) seconds, and the approximate time (rounded to the nearest hundredth) is \( 3.78 \) seconds. For the exact time part (as per the first part of the question), the answer is \( \frac{10\sqrt{7}}{7} \) (or approximately \( 3.78 \) when rounded, but the exact form is \( \frac{10\sqrt{7}}{7} \)).