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

Answer:

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

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.

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

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.