Question

jouri is passing her key to a friend. from her balcony, she drops the key from a height of 150 feet. unfortunately, her friend did not catch the key. find the time when the key is 25 feet above the ground. round your answer to the nearest tenth if necessary. use the following formulas:
$h = -\frac{1}{2}gt^2 + v_0t + h_0$
variables: $h$ = height of the object, $t$ = time in motion
constants: $g$ = gravity (either $32\\ \text{ft/s}^2$ or $9.8\\ \text{m/s}^2$), $v_0$ = initial velocity, $h_0$ = initial height
show your work here

Explanation:

Step1: Substitute the values into the formula

Since the key is dropped, \(v_0 = 0\) (initial velocity), \(h_0=150\) (initial height), \(g = 32\) (gravity in feet - per - second - squared), and \(h = 25\) (desired height).
The formula \(h=-\frac{1}{2}gt^{2}+v_0t + h_0\) becomes \(25=-\frac{1}{2}\times32t^{2}+0\times t + 150\).
Simplify the equation: \(25=-16t^{2}+150\).

Step2: Rearrange the equation to solve for \(t^{2}\)

Subtract 150 from both sides: \(25 - 150=-16t^{2}\).
\(-125=-16t^{2}\).
Then \(t^{2}=\frac{125}{16}\).

Step3: Solve for \(t\)

Take the square root of both sides: \(t=\sqrt{\frac{125}{16}}\).
\(t=\frac{\sqrt{125}}{4}=\frac{5\sqrt{5}}{4}\approx\frac{5\times2.236}{4}\).
\(t\approx2.8\)

Answer:

\(t\approx2.8\) seconds