A clever inventor has created a device that can launch water balloons with an initial speed of 85.0 m/s. Her goal is to pass a b

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.