Question

josh is competing at a hot - air balloon festival. from a balloon directly over a target, he drops a marker from a height of 60 meters. find the time when the marker 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

Explanation:

Step1: Substitute the values into the formula

Given \(h = 5\) meters (height above ground), \(g=9.8\ m/s^{2}\), \(v_{0} = 0\) (dropped, so initial velocity is \(0\)), \(h_{0}=60\) meters.
Substitute into \(h =-\frac{1}{2}gt^{2}+v_{0}t + h_{0}\):
\(5=-\frac{1}{2}(9.8)t^{2}+0\times t + 60\)

Step2: Simplify the equation

\(5=-4.9t^{2}+60\)
Subtract \(60\) from both sides:
\(5 - 60=-4.9t^{2}\)
\(- 55=-4.9t^{2}\)
Multiply both sides by \(-1\): \(55 = 4.9t^{2}\)
Then \(t^{2}=\frac{55}{4.9}\)

Step3: Solve for \(t\)

\(t=\sqrt{\frac{55}{4.9}}\)
\(t=\sqrt{11.22449}\)
\(t\approx3.3\)

Answer:

\(t\approx3.3\) seconds