when an object is dropped from the top of a 75 ft tall building, the object will be h feet above the ground after t seconds,where t = $\frac{1}{4}\sqrt{75-h,}h\le75$‍14 √75−h,h≤75. How far above the ground will the object be after 1 s? Later verify answers.

6 years ago

Answered By Gayathry C

Find the answer by using the given formula, using t=1.

1=$\frac{1}{4}\sqrt{75-h}$14 √75−h‍ 

4= $\sqrt{75-h}$‍√75−h 

16 =75-h

h=75-16

h=59 ft

 

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