Question

lenn is passing his key to a friend. from his balcony, he drops the key from a height of 45 meters. unfortunately, his friend did not catch the key. find the time when the key is 5 meters 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 $32$ ft/$s^2$ or $9.8$ m/$s^2$), $v_0$ = initial velocity, $h_0$ = initial height
show your work here
seconds

Explanation:

Step1: Substitute values into formula

Given \(h = 5\), \(g = 9.8\), \(v_0=0\) (dropped, so initial velocity \(0\)), \(h_0 = 45\).
Substitute into \(h=-\frac{1}{2}gt^{2}+v_0t + h_0\):
\(5=-\frac{1}{2}(9.8)t^{2}+0\times t + 45\).
Simplify to \(5=-4.9t^{2}+45\).

Step2: Rearrange equation

Subtract \(45\) from both sides:
\(5 - 45=-4.9t^{2}\).
\(- 40=-4.9t^{2}\).
Divide both sides by \(-4.9\): \(t^{2}=\frac{40}{4.9}\approx8.163\).

Step3: Solve for \(t\)

Take square root of both sides: \(t=\sqrt{8.163}\approx2.9\) (since \(t>0\) for time).

Answer:

\(2.9\)