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 Andrew T
What we are trying to find is what h equals to when t=1. Therefore, substitute t=1 into the equation.
$1=\frac{1}{4}\sqrt{75-h}$1=14√75−h (multiply 4 on both sides)
$4=\sqrt{75-h}$4=√75−h (square both sides to eliminate the root)
$16=75-h$16=75−h (basic algebra from here)
$h=59$h=59 (hight of the object after 1s is 59m)
Verify: simply substitute h=59. If your result is t=1, then you know you got the right answer.
6 years ago
Answered By Andrew T
What we are trying to find is what h equals to when t=1. Therefore, substitute t=1 into the equation.
$1=\frac{1}{4}\sqrt{75-h}$1=14 √75−h (multiply 4 on both sides)
$4=\sqrt{75-h}$4=√75−h (square both sides to eliminate the root)
$16=75-h$16=75−h (basic algebra from here)
$h=59$h=59 (hight of the object after 1s is 59m)
Verify: simply substitute h=59. If your result is t=1, then you know you got the right answer.
$t=\frac{1}{4}\sqrt{75-59}$t=14 √75−59
$t=\frac{1}{4}\sqrt{16}$t=14 √16
$t=\frac{1}{4}\left(4\right)$t=14 (4)
$t=1$t=1