Average velocity can be found by taking the total distance traveled in the two part motion and dividing by the total time of the two part motion as shown by the equation for velocity:

$\displaystyle v=\frac{\Delta x}{\Delta t}$

Where $\displaystyle v$ is velocity, $\displaystyle x$ is displacement, and $\displaystyle t$ is time.

Our average velocity then would be:

$\displaystyle v_a_v= \frac{x_1+x_2}{t_1+t_2}$

While we were not given the displacements for either part, we can solve for them by rearranging the velocity equation:

$\displaystyle x_1=(v_1)(t_1), x_2=(v_2)(t_2)$

Substituting for the unknown displacements with equivalent terms comprised of the known velocities and times we find that the average velocity is:

 $\displaystyle \frac{(v_1t_1)+(v_2t_2)}{t_1+t_2}$

In a traditional coordinate system: left and down are considered negative directions while up and right are considered positive directions. The question states that the object is moving down, so regardless of its acceleration, its velocity is negative at the time described.

In a traditional coordinate system: left and down are considered negative directions while up and right are considered positive directions. Though the object is moving down (a negative direction) it is stated that it is slowing down, which means it is accelerating upwards in a positive direction.

In a traditional coordinate system: left and down are considered negative directions while up and right are considered positive directions. Though the car is said to be speeding up, it is speeding up while moving in a negative direction. Therefor, the car's velocity is becoming more and more negative, meaning it has a negative acceleration.

With only the drop height and the time we can use the kinematic equation:

$\displaystyle x=v_ot+\frac{1}{2}at^2$

We assume the initial velocity is $\displaystyle 0\frac{m}{s}$ because the ball is dropped, plug in the height for $\displaystyle x$, the time from drop to hitting the ground for $\displaystyle t$, and then the only unknown variable is $\displaystyle a$ which will be the experimental value Earth's gravitational acceleration.

We use the kinematic equation:

$\displaystyle x=v_ot+\frac{1}{2}at^2$

Plugging in $\displaystyle 8.0m$ for $\displaystyle x$, $\displaystyle 0\frac{m}{s}$ for $\displaystyle v_o$ because it was dropped from rest, and $\displaystyle 1s$ for $\displaystyle t$ we have:

$\displaystyle -8m=(0\frac{m}{s})*(1s)+\frac{1}{2}a*(1s)^2$

$\displaystyle -8m=0+\frac{1}{2}a(1s^2)$

$\displaystyle -16\frac{m}{s^2}=a$

In the kinematic equation:

$\displaystyle x=v_ot+\frac{1}{2}at^2$

When the first term right of the equals sign goes to zero because the penny was dropped from rest, we see that the distance in a constant acceleration $\displaystyle (x)$ is related to the time of acceleration $\displaystyle (t)$ via:

$\displaystyle x\propto t^2$

because the fraction and acceleration are constant.

So when $\displaystyle t$ is tripled from $\displaystyle 2s$ to $\displaystyle 6s$ ($\displaystyle t_1*3=t_2$), the distance increases ninefold ($\displaystyle x_1*9=x_2$)

Horizontal velocity does not affect gravity's force on an object so both the bullet and the ball are accelerated downward at the same gravitational acceleration and they will hit the ground at the same time.

So for this first we have to set up a relation between John (1) and Dave's (2) position. Keeping in mind that Dave is traveling the opposite distance of John.

$\displaystyle v_{1}t - x = -v_{2}t$

Where x is the total distance between the two points. Then we solve it for time and plug it in to John's speed to find his distance after the elapsed time.

$\displaystyle v_{1}t + v_{2}t = x$

$\displaystyle t = \frac{x}{v_{1}+v_{2}}$

$\displaystyle v_{1}t = x_{1}$

$\displaystyle \frac{v_{1}x}{v_{1}+v_{2}}=x_{1}$

where $\displaystyle x_{1}$ is John's total distance traveled. Then we plug in our values

$\displaystyle \frac{35*60}{35 + 50} = 24.7 mi = x_{1}$