Question

lise is passing her key to a friend. from her balcony, she drops the key from a height of 45 meters. unfortunately, her 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 the values into the formula

Given \(h = 5\) meters, \(v_0=0\) (since the key is dropped), \(h_0 = 45\) meters and \(g = 9.8\ m/s^{2}\).
Substitute into \(h=-\frac{1}{2}gt^{2}+v_0t + h_0\), we get \(5=-\frac{1}{2}\times9.8t^{2}+0\times t + 45\).
Simplify the equation: \(5=-4.9t^{2}+45\).

Step2: Rearrange the equation

Subtract 45 from both sides: \(5 - 45=-4.9t^{2}\).
So, \(- 40=-4.9t^{2}\).
Then \(t^{2}=\frac{40}{4.9}\).

Step3: Solve for \(t\)

\(t=\sqrt{\frac{40}{4.9}}\approx\sqrt{8.163}\approx2.9\)

Answer:

\(2.9\)