Question

1 practice problems on kinematics

1. an object is thrown from the ground vertically upwards with an initial speed $v_0$ and returns to its starting point after a time of 6.0 s. neglecting air resistance, how high above the ground does it go?
2. at what height above the ground is the object in problem 1 traveling at one - third of its initial speed $v_0$?
3. a car is traveling at 20 m/s towards a stop sign and begins to brake when it is 30 m from the sign. with what acceleration must the car brake in order to stop exactly at the sign (this is the minimum braking that must be applied to stop before going past the sign)?
4. how long does it take the car in problem 3 to come to a complete stop if it brakes so that it stops exactly at the sign?

Explanation:

Step1: Find the initial velocity for problem 1

The time - of - flight formula for vertical motion is $t = \frac{2v_0}{g}$, where $t = 6.0\ s$ and $g=9.8\ m/s^{2}$. Rearranging for $v_0$ gives $v_0=\frac{gt}{2}$.
$v_0=\frac{9.8\times6.0}{2}=29.4\ m/s$
The maximum - height formula is $h=\frac{v_0^{2}}{2g}$. Substituting $v_0 = 29.4\ m/s$ and $g = 9.8\ m/s^{2}$ gives $h=\frac{(29.4)^{2}}{2\times9.8}=44.1\ m$

Step2: Solve problem 2

We know the kinematic equation $v^{2}=v_0^{2}-2gh$. Given $v=\frac{v_0}{3}$, we substitute into the equation: $(\frac{v_0}{3})^{2}=v_0^{2}-2gh$.
Expanding, $\frac{v_0^{2}}{9}=v_0^{2}-2gh$. Rearranging for $h$: $2gh = v_0^{2}-\frac{v_0^{2}}{9}=\frac{8v_0^{2}}{9}$. So $h=\frac{4v_0^{2}}{9g}$. Since $v_0 = 29.4\ m/s$ and $g = 9.8\ m/s^{2}$, $h=\frac{4\times(29.4)^{2}}{9\times9.8}=39.2\ m$

Step3: Solve problem 3

The kinematic equation $v^{2}=v_0^{2}+2ax$ is used. Here, $v = 0\ m/s$, $v_0=20\ m/s$ and $x = 30\ m$. Rearranging for $a$ gives $a=\frac{v^{2}-v_0^{2}}{2x}$.
$a=\frac{0 - 20^{2}}{2\times30}=-\frac{20}{3}\approx - 6.67\ m/s^{2}$

Step4: Solve problem 4

The kinematic equation $v=v_0+at$ is used. Since $v = 0\ m/s$, $v_0 = 20\ m/s$ and $a=-\frac{20}{3}\ m/s^{2}$, rearranging for $t$ gives $t=\frac{v - v_0}{a}$.
$t=\frac{0 - 20}{-\frac{20}{3}}=3\ s$

Answer:

1. $44.1\ m$
2. $39.2\ m$
3. $-6.67\ m/s^{2}$
4. $3\ s$