. A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00m above the ground. At the same instant the balloon is released; a second balloon is thrown straight down at 2.00 m/s from the same height. Determine which balloon hits the ground first and how much sooner it hits the ground than the other balloon

Explanation:

Height of building

H = 6m

Horizontal speed of first balloon

U1x = 2m/s

Second ballot is thrown straight downward at a speed of

U2y = 2m/s

Time each gallon hits the ground

Balloon 1.

Using equation of free fall

H = Uoy•t + ½gt²

Uox = 0 since the body does not have vertical component of velocity

6 = ½ × 9.8t²

6 = 4.9t²

t² = 6 / 4.9

t² = 1.224

t = √1.224

t = 1.11 seconds

For second balloon

H = Uoy•t + ½gt²

6 = 2t + ½ × 9.8t²

6 = 2t + 4.9t²

4.9t² + 2t —6 = 0

Using formula method to solve the quadratic equation

Check attachment

From the solution we see that,

t = 0.9211 and t = -1.329

We will discard the negative value of time since time can't be negative here

So the second balloon get to the ground after t ≈ 0.92 seconds

Conclusion

The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..