Question

while riding a roller coaster, a girl drops an object. the roller coaster was rising vertically at a velocity of 11.0m/s and was 5.00m above the ground when the object was dropped. how long does it take to reach the ground

Answer 1

Approximately
`0.388\; {\rm s}`, assuming that air resistance is negligible and that
`g = 9.81\; {\rm m\cdot s^(-2)}`.

Step-by-step explanation:

Initial vertical velocity of the object:
`u = 11.0\; {\rm m\cdot s^(-1)}`, upwards (same as that of the rollercoaster.)

Initial height of the object:
`h_(0) = 5.00\; {\rm m}`.

If air resistance is negligible, this object will accelerate downwards at a constant
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}`. Note that
`a` is negative since the object is accelerating downwards.

The SUVAT equation
`h = (1/2)\, a\, t^(2) + u\, t + h_(0)` gives the height
`h` of this object at time
`t`. Note that while the initial height is
`5.00\; {\rm m}`,
`h = 0` when the object reaches the ground.

Since acceleration
`a`, initial velocity
`u`, and initial height
`h_(0)` are all given, setting
`h\!` to
`0` and solving for
`t` will give the time it takes for this object to reach the ground:

`(1/2)\, a\, t^(2) + u\, t + h_(0) = 0`.

`(1/2)\, (-9.81) \, t^(2) + 11.0\, t + 5.00 = 0`.

`(-4.905)\, t^(2) + 11.0\, t + 5.00 = 0`.

Solve this equation for
`t\!` using the quadratic formula. Note that
`t > 0` since
`t` denotes the amount of time required for the object to reach the ground.

`\begin{aligned} t &= \frac{-11.0 + \sqrt{11.0^(2) - 4 * (-4.905) * 5.00}}{2* (-4.905)} \\ &\approx 0.388\; {\rm s}\end{aligned}`.

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
`0.388\; {\rm s}` for this object to reach the ground.