Question

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 balloon through a small hoop mounted on the observation platform at the top of the Eiffel Tower, 276 m above the ground. If the balloon is to pass through the hoop at the peak of its flight, at what angle above horizontal should she launch the balloon? Please explain step by step

Answer 1

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.