Answer:
Approximately [tex]0.388\; {\rm s}[/tex], assuming that air resistance is negligible and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
Initial vertical velocity of the object: [tex]u = 11.0\; {\rm m\cdot s^{-1}}[/tex], upwards (same as that of the rollercoaster.)
Initial height of the object: [tex]h_{0} = 5.00\; {\rm m}[/tex].
If air resistance is negligible, this object will accelerate downwards at a constant [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex]. Note that [tex]a[/tex] is negative since the object is accelerating downwards.
The SUVAT equation [tex]h = (1/2)\, a\, t^{2} + u\, t + h_{0}[/tex] gives the height [tex]h[/tex] of this object at time [tex]t[/tex]. Note that while the initial height is [tex]5.00\; {\rm m}[/tex], [tex]h = 0[/tex] when the object reaches the ground.
Since acceleration [tex]a[/tex], initial velocity [tex]u[/tex], and initial height [tex]h_{0}[/tex] are all given, setting [tex]h\![/tex] to [tex]0[/tex] and solving for [tex]t[/tex] will give the time it takes for this object to reach the ground:
[tex](1/2)\, a\, t^{2} + u\, t + h_{0} = 0[/tex].
[tex](1/2)\, (-9.81) \, t^{2} + 11.0\, t + 5.00 = 0[/tex].
[tex](-4.905)\, t^{2} + 11.0\, t + 5.00 = 0[/tex].
Solve this equation for [tex]t\![/tex] using the quadratic formula. Note that [tex]t > 0[/tex] since [tex]t[/tex] denotes the amount of time required for the object to reach the ground.
[tex]\begin{aligned} t &= \frac{-11.0 + \sqrt{11.0^{2} - 4 \times (-4.905) \times 5.00}}{2\times (-4.905)} \\ &\approx 0.388\; {\rm s}\end{aligned}[/tex].
The other root of this quadratic equation is negative and isn't a valid solution to the question.
In other words, it will take approximately [tex]0.388\; {\rm s}[/tex] for this object to reach the ground.
