A gas has an initial volume of 168 cm3 at a temperature of 255 K and a pressure of 1.6 atm. The pressure of the gas decreases to
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Answer:
She should launch the balloon at an angle of 59.9° above the horizontal.
Explanation:
Please, see the attached figure for a graphical description of the problem.
The position and velocity vectors of the water balloon at time "t" can be obtained using the following equations:
r = (x0 + v0 · t · cos θ, y0 + v0 · t · sin θ + 1/2 · g · t²)
v = (v0 · cos θ, v0 · sin θ + g · t)
Where:
r = position vector at time "t".
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
θ = launching angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
v = velocity vector at time "t".
Let´s place the origin of the frame of reference at the launching point so that x0 and y0 = 0.
At the maximum height (276 m), the vector velocity of the balloon is horizontal (see v1 in the figure). That means that the y-component of the velocity vector is 0. Then, using the equation of the y-component of the velocity vector, we can write:
At maximum height:
vy = v0 · sin θ + g · t
0 = v0 · sin θ + g · t
We also know that at maximum height, the y-component of the position vector is 276 m (see r1y in the figure). Then:
At maximum height:
y = y0 + v0 · t · sin θ + 1/2 · g · t²
276 m = y0 + v0 · t · sin θ + 1/2 · g · t²
So, we have two equations with two unknowns (θ and t):
276 m = y0 + v0 · t · sin θ + 1/2 · g · t²
0 = v0 · sin θ + g · t
To solve the system of equations, let´s take the equation of the y-component of the velocity and solve it for sin θ. Then, we will replace sin θ in the equation of the y-component of the position to obtain the time and finally obtain θ:
0 = v0 · sin θ + g · t
0 = 85.0 m/s · sin θ - 9.81 m/s² · t
9.81 m/s² · t / 85.0 m/s = sin θ
Replacing sin θ in the equation of the vertical component of the position:
276 m = y0 + v0 · t · sin θ + 1/2 · g · t² (y0 = 0)
276 m = 85.0 m/s · t · (9.81 m/s² · t /85. 0 m/s) - 1/2 · 9.81 m/s² · t²
276 m = 9.81 m/s² · t² - 1/2 · 9.81 m/s² · t²
276 m = 1/2 · 9.81 m/s² · t²
276 m / ( 1/2 · 9.81 m/s²) = t²
t = 7.50 s
Now, we can calculate the angle θ using the equation obtained above:
9.81 m/s² · t / 85.0 m/s = sin θ
9.81 m/s² · 7.50 s / 85.0 m/s = sin θ
θ = 59.9°
She should launch the balloon at an angle of 59.9° above the horizontal.
