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choli [55]
3 years ago
13

A person throws a stone from the corner edge of a building. The stone's initial velocity is 28.0 m/s directed at 43.0° above the

horizontal. The stone hits tip of a structure which is 5 m tall, 6.1 s after being thrown. Ignore air resistance.
(a) What is the height of the building?
(b) How far from the foot of the building does the stone land (only horizontal distance until the foot of the structure)?
(c) How fast is the stone moving just before it hits the ground?
(d) Please write down the acceleration vector in the unit vector notation for the stone at the instant it hits the ground.
Physics
1 answer:
Naya [18.7K]3 years ago
4 0

The stone's acceleration, velocity, and position vectors at time t are

\mathbf a(t)=-g\,\mathbf j

\mathbf v(t)=v_{i,x}\,\mathbf i+\left(v_{i,y}-gt\right)\,\mathbf j

\mathbf r(t)=v_{i,x}t\,\mathbf i+\left(y_i+v_{i,y}t-\dfrac g2t^2\right)\,\mathbf j

where

g=9.80\dfrac{\rm m}{\mathrm s^2}

v_{i,x}=\left(28.0\dfrac{\rm m}{\rm s}\right)\cos43.0^\circ\approx20.478\dfrac{\rm m}{\rm s}

v_{i,y}=\left(28.0\dfrac{\rm m}{\rm s}\right)\sin43.0^\circ\approx19.096\dfrac{\rm m}{\rm s}

and y_i is the height of the building and initial height of the rock.

(a) After 6.1 s, the stone has a height of 5 m. Set the vertical component (\mathbf j) of the position vector to 5 m and solve for y_i:

5\,\mathrm m=y_i+\left(19.096\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.1\,\mathrm s)^2

\implies\boxed{y_i\approx70.8\,\mathrm m}

(b) Evaluate the horizontal component (\mathbf i) of the position vector when t=6.1\,\mathrm s:

\left(20.478\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)\approx\boxed{124.92\,\mathrm m}

(c) The rock's velocity vector has a constant horizontal component, so that

v_{f,x}=v_{i,x}\approx20.478\dfrac{\rm m}{\rm s}

where v_{f,x}

For the vertical component, recall the formula,

{v_{f,y}}^2-{v_{i,y}}^2=2a\Delta y

where v_{i,y} and v_{f,y} are the initial and final velocities, a is the acceleration, and \Delta y is the change in height.

When the rock hits the ground, it will have height y_f=0. It's thrown from a height of y_i, so \Delta y=-y_i. The rock is effectively in freefall, so a=-g. Solve for v_{f,y}:

{v_{f,y}}^2-\left(19.096\dfrac{\rm m}{\rm s}\right)^2=2(-g)(-124.92\,\mathrm m)

\implies v_{f,y}\approx-53.039\dfrac{\rm m}{\rm s}

(where we took the negative square root because we know that v_{f,y} points in the downward direction)

So at the moment the rock hits the ground, its velocity vector is

\mathbf v_f=\left(20.478\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(-53.039\dfrac{\rm m}{\rm s}\right)\,\mathbf j

which has a magnitude of

\|\mathbf v_f\|=\sqrt{\left(20.478\dfrac{\rm m}{\rm s}\right)^2+\left(-53.039\dfrac{\rm m}{\rm s}\right)^2}\approx\boxed{56.855\dfrac{\rm m}{\rm s}}

(d) The acceleration vector stays constant throughout, so

\mathbf a(t)=\boxed{-g\,\mathbf j}

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A sinewave has a period (duration of one cycle) of 645 μs (microseconds). What is the corresponding frequency of this sinewave,
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The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.

<u>Given the following data:</u>

  • Period = 645 μs

Note: μs represents microseconds.

<u>Conversion:</u>

1 μs = 1 × 10^-6 seconds

645 μs = 645 × 10^-6 seconds

To find corresponding frequency of this sinewave, in kHz;

Mathematically, the frequency of a waveform is calculated by using the formula;

Frequency = \frac{1}{Period}

Substituting the value into the formula, we have;

Frequency = \frac{1}{645 * 10^-6}

Frequency = 1550.39 Hz

Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);

<u>Conversion:</u>

1 hertz = 0.001 kilohertz

1550.39 hertz = X kilohertz

Cross-multiplying, we have;

X = 0.001 × 1550.39

X = 155039 kHz

To 3 significant figures;

<em>Frequency = 155 kHz</em>

Therefore, the corresponding frequency of this sinewave, in kHz is 155.

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Respuesta:

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Explicación:

Paso 1: Información provista

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Paso 2: Convertir la velocidad del coyote de km/h a m/s

Vamos a usar los siguientes factores de conversion:

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65 km/h × (1000 m/1 km) × (1 h/3600 s) = 18 m/s

Paso 3: Calcular la energía cinética del coyote (K)

Usarémos la siguiente fórmula.

K = 1/2 × m × v²

K = 1/2 × 30 kg × (18 m/s)² = 4.9 × 10³ J

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