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Anarel [89]
3 years ago
5

A person throws a pumpkin at a horizontal speed of 4.0 — off a cliff. The pumpkin travels 9.5 m horizontally

Physics
1 answer:
emmainna [20.7K]3 years ago
7 0

Complete Question

A person throws a pumpkin at a horizontal speed of   4.0 m/s off a cliff. The pumpkin travels 9.5m horizontally before it hits the ground. We can ignore air resistance.What is the pumpkin's vertical displacement during the throw? What is the pumpkin's vertical velocity when it hits the ground?

Answer:

The  pumpkin's vertical displacement  is  H = 27 .67 \ m

The  pumpkin's vertical velocity when it hits the ground is  v_v__{f}} = 23.298 \  m/s

Explanation:

From the question we are told that

   The  horizontal speed is  v_h  =  4 m/s

    The horizontal distance traveled is  d =  9.5 \ m

The horizontal distance traveled is mathematically represented as

           S =  v_h * t

Where t is the time taken

substituting values

          9.5 =  4 * t

   =>     t =  \frac{9.5}{4}

            t = 2.38 \ sec

Now the vertical displacement is mathematically represented as

        H  =  v_v t  +  \frac{1}{2} a_v t^2

now the vertical velocity before the throw is  zero

    So

          H =  0 +  \frac{1}{2} (9.8) * (2.375)^2

          H = 27 .67 \ m

Now the final vertical velocity  is mathematically represented as

          v_v__{f}} =  v_v + at

  substituting values

             v_v__{f}} =  0 + (9.8)* (2.375)

            v_v__{f}} = 23.298 \  m/s

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Answer:

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Explanation:

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Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

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Answer:

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Now let's use the magnification ratio

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