You decide to fly over to Fred Meyer’s using your jet pack but during your vertical ascent you drop your wallet. If your wallet is dropped when you are ascending at a constant speed of 6 m/s and the wallet is released with the same upward velocity of 6 m/s when t = 0. Determine the speed of your wallet when it hits the ground at t = 8s. Also determine the altitude of your jet pack.

Here we only need to look at the vertical problem, so first, let's look at the forces acting vertically on the wallet.

When the wallet starts to fall, the only force acting on it will be the gravitational force (where we are ignoring the effects of air friction).

Then the acceleration of the wallet will be equal to the gravitational acceleration, g = 9.8m/s^2

Then we can write:

a(t) = (-9.8m/s^2)

Where the negative sign is because this acceleration is downwards.

To get the vertical velocity equation of the wallet we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the constant of integration, and in this case is the initial velocity of the wallet, which we know is equal to 6m/s, then the velocity equation is:

v(t) = (-9.8 m/s^2)*t + 6m/s

To get the position equation we need to integrate over time again, we get:

p(t) = (1/2)*(-9.8 m/s^2)*t^2 + (6m/s)*t + p0

Where p0 is the initial vertical position, in this case, is the height at which the wallet is dropped, which is also the altitude of your jet pack when the wallet falls.

Now we want to know two things:

Determine the speed of your wallet when it hits the ground at t = 8s

Here we just need to evaluate the velocity equation in t = 8s.

v(8s) =  (-9.8 m/s^2)*8s + 6m/s = -72.4 m/s

We also want to determine the altitude of the jet pack (when the wallet drops).

To find this, we can use the fact that the wall hits the ground at t = 8s.

The wallet hits the ground when it's vertical position is equal to zero, then:

p(8s) = 0m = (1/2)*(-9.8 m/s^2)*(8s)^2 + (6m/s)*8s + p0

Now we can solve this for p0.

0m = (1/2)*(-9.8 m/s^2)*(8s)^2 + (6m/s)*8s + p0

(1/2)*(9.8 m/s^2)*(8s)^2 -  (6m/s)*8s = p0

265.6m = p0

This means that the altitude of the jet pack when the wallet drops is 265.6m

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?