Answer:
She must launch the balloon at a 59.9° angle above the horizontal plane.
Explanation:
Refer to the accompanying diagram for a visual explanation of the scenario.
The position and velocity vectors of the balloon at time "t" can be computed using these formulas:
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 = launch angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.81 m/s² if upward is treated as positive).
v = velocity vector at time "t".
We can establish the origin of the reference frame at the launch point, making x0 and y0 = 0.
At the maximum altitude (276 m), the balloon's velocity vector is horizontal (refer to v1 in the diagram). Thus, the y-component of the velocity vector is zero. We can express this using the velocity vector's y-component equation:
At maximum altitude:
vy = v0 · sin θ + g · t
0 = v0 · sin θ + g · t
Also, at maximum height, the y-component of the position vector is 276 m (see r1y in the diagram). Therefore:
At peak height:
y = y0 + v0 · t · sin θ + 1/2 · g · t²
276 m = y0 + v0 · t · sin θ + 1/2 · g · t²
We have two equations involving two unknowns (θ and t):
276 m = y0 + v0 · t · sin θ + 1/2 · g · t²
0 = v0 · sin θ + g · t
To solve for the variables, we can rearrange the velocity equation for sin θ and substitute it back into the position equation to find t and eventually 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
Now, substituting sin θ in the position equation:
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
With t known, we can find the angle θ using the previously derived formula:
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 a 59.9° angle above the horizontal.
Answer:
All three pendulums will have the same angular frequencies.
Explanation:
For a simple pendulum, the time period using the approximation is expressed as:
The angular frequency is defined as
Since the angular frequency remains unaffected by the initial angle (valid strictly for small angle approximations), we deduce that the angular frequencies of the three pendulums are identical.