4. An object is thrown from from the ground upward with an initial speed of 3.75 m/s. How long will the object be in the air before it lands on the ground?​

Explanation:

There's an easy way to answer this and then an easier way. I'll do both since I'm not sure what you're doing this for: physics or calculus. Calculus is the easier way, btw.

Going with the physics version first, here's what we know:

a = -9.8 m/s/s

v₀ = 3.75 m/s

t = ??

That's not a whole lot...at least not enough to directly solve the problem. What we have to remember here is that at the max height of a parabolic path, the final velocity is 0. So we can add that to our info:

v = 0 m/s. Use the one-dimensional equation that utilizes all that info and allows us to solve for time:

v = v₀ +at and filling in:

0 = 3.75 + (-9.8)t and

-3.75 = -9.8t so

t = .38 seconds. This is how long it takes to get to its max height. Another thing we need to remember (which is why calculus is so much easier!) is that at the halfway point of a parabolic path (the max height), the object has traveled half the time it takes to make the whole trip. In other words, if .38 is how long it takes to go halfway, then 2(.38) is how long the whole trip takes:

2(.38) = .76 seconds. Now onto the calculus way:

The position function is

The first derivative of this is the velocity function and, knowing that when the velocity is 0, the time is halfway gone, we will find the velocity function and then set it equal to 0 and solve for t:

v(t) = -9.8t + 3.75 and

0 = -9.8t + 3.75 and

-3.75 = -9.8t so

t = ,38 and multiply that by 2 to find the time the whole trip took:

2(.38) = .76 seconds.

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j