# A rock is thrown upward with an initial velocity of 14m/s. The motion of the rock can be modelled by the equation h(t) + -4.9t^2 +14t ?

## to the nearest hundredth of a second, for what period of time is the rock's altitude greater than 6m? for how long is the rock's altitude greater than 6m? answer to the nearest hundredth of a second

Apr 26, 2018

Yes.

#### Explanation:

Here is how to solve a projectile motion with garanteed sucess.
1) Set your vertical axis up.
2) Project all vectors on that axis
3) Write Newton's law

In detail now.

The force of gravity is down. So gravity projected onto the vertical axis is -g where g is positive and equal to 9.81 m/s^2.

Now we write Newton's law in vector form

Sum of forces = mass times acceleration.

$\vec{F} = m \setminus \vec{a}$
Project on upward axis
$- g = \setminus \frac{\mathrm{dv}}{\mathrm{dt}}$

Multiply by dt
$- g \mathrm{dt} = \mathrm{dv}$

Integrate once
$- g \left(t - {t}_{0}\right) = v \left(t\right) - v \left({t}_{0}\right)$

In this expression, v(t) varies, v(t_0) is a number. Call it v_0

$- g \left(t - {t}_{0}\right) = v \left(t\right) - {v}_{0}$

Now replace v(t) by $\frac{\mathrm{dh}}{\mathrm{dt}}$

$- g \left(t - {t}_{0}\right) = \setminus \frac{\mathrm{dh}}{\mathrm{dt}} - {v}_{0}$

Mutliply by $\mathrm{dt}$

$- g \cdot t \mathrm{dt} + g \cdot {t}_{0} \mathrm{dt} + {v}_{0} = \mathrm{dh}$

Integrate

$- g \left({t}^{2} / 2\right) - g \cdot {t}_{0}^{2} / 2 + g \cdot {t}_{0} \cdot \left(t - {t}_{0}\right) + {v}_{0} \left(t - {t}_{0}\right) = h \left(t\right) - h \left({t}_{0}\right)$

Now we have two terms of the form $g \cdot {t}_{0}^{2}$

Let's expand.

$- g \left({t}^{2} / 2\right) + g \cdot {t}_{0}^{2} / 2 + g \cdot {t}_{0} \cdot t - g \cdot {t}_{0}^{2} + \cdot {v}_{0} \left(t - {t}_{0}\right) = h \left(t\right) - h \left({t}_{0}\right)$

$h \left(t\right) - h \left({t}_{0}\right) = - \frac{1}{2} g {\left(t - {t}_{0}\right)}^{2} + {v}_{0} \left(t - {t}_{0}\right)$

In this problem, ${t}_{0} = 0$ and ${v}_{0} = 14 \frac{m}{s}$

We obtain
$h \left(t\right) = - 4.9 {t}^{2} + 14 t$

What must be the value of t such that h is equal to 6?
We have a quadratic equation for h
$4.9 {t}^{2} - 14 t - 6 = 0$

whose positive root is $t = \setminus \frac{7 + \sqrt{{7}^{2} + 29.4}}{4.9} = 3.23 s$

graph{-4.9x^2 + 14x [0, 4, 0, 11]}

The maximum height satisfies

$9.8 t - 14 = 0$

$t = \frac{7}{4.9}$

The rock reaches a height of 6 meters at $t = \frac{7}{4.9} + \frac{\sqrt{49 + 29.4}}{4.9}$. The time spent above h=6 is the time spend above h=6 while going up plus the time spend going down. These are equal. Since t = t(h=6) + sqrt(49+29.4)/4.9#.

Hence
time spent since h= 6 is $2 \cdot \frac{\sqrt{49 + 29.4}}{4.9} = 6.46$