Answer:
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.
Answer:

Explanation:



Electron information needed to solve the question:






![E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B9.11x10%7B-31%7Dkg%2A3.0x10%5E%7B12%7Dm%2Fs%5E2%7D%7B-1.6x10%7B-19%7DC%7D-%5B%2819.0x10%5E3mj%2B18.0x10%5E3m%29xi%28400x10%5E%7B-6%7DT%29%5D)
![E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C](https://tex.z-dn.net/?f=E%3D-i17.08N%2FC-%5B7.6%28-k%29%2B7.2%28j%29%5DN%2FC)

Answer:
+2m/s
Explanation:
average velocity = displacement traveled / total time taken
= +12m/ 6s
= +2 m/s
Answer:
The function that describe the motion in the time
y (t) = 0.28m * sin ( 36.025 * t)
Explanation:
The angular frequency of oscillation of the spring
w = √k/m
w = √305 N/m / 0.235 kg
w = 36.025 rad / s
To determine the function of the motion knowing as a motion oscillation in a amplitude a frequency
y(t) = A * sin (w t )
So
A = 28.0 cm * 1 m / 100 cm = 0.28 m
So replacing to determine the function of the motion in the time
y (t) = A sin (w t)
y (t) = 0.28m * sin ( 36.025 * t)