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nordsb [41]
3 years ago
8

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

alloon 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

Physics
1 answer:
seropon [69]3 years ago
6 0

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.

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Current Flow and Ohm's Law

Ohm's law is the most important, basic law of electricity. It defines the relationship between the three fundamental electrical quantities: current, voltage, and resistance. When a voltage is applied to a circuit containing only resistive elements (i.e. no coils), current flows according to Ohm's Law, which is shown below.

I = V / R 

Where: 

I =

Electrical Current (Amperes)

V =

Voltage (Voltage)

R =

Resistance (Ohms)

    

Ohm's law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). Therefore, if the voltage is increased, the current will increase provided the resistance of the circuit does not change. Similarly, increasing the resistance of the circuit will lower the current flow if the voltage is not changed. The formula can be reorganized so that the relationship can easily be seen for all of the three variables.

The Java applet below allows the user to vary each of these three parameters in Ohm's Law and see the effect on the other two parameters. Values may be input into the dialog boxes, or the resistance and voltage may also be varied by moving the arrows in the applet. Current and voltage are shown as they would be displayed on an oscilloscope with the X-axis being time and the Y-axis being the amplitude of the current or voltage. Ohm's Law is valid for both direct current (DC) and alternating current (AC). Note that in AC circuits consisting of purely resistive elements, the current and voltage are always in phase with each other.

Exercise: Use the interactive applet below to investigate the relationship of the variables in Ohm's law. Vary the voltage in the circuit by clicking and dragging the head of the arrow, which is marked with the V. The resistance in the circuit can be increased by dragging the arrow head under the variable resister, which is marked R. Please note that the vertical scale of the oscilloscope screen automatically adjusts to reflect the value of the current.

See what happens to the voltage and current as the resistance in the circuit is increased. What happens if there is not enough resistance in a circuit? If the resistance is increased, what must happen in order to maintain the same level of current flow?


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