In a circus act, a 78 kg clown is shot from a cannon with an initial velocity of 19 m/s at some unknown angle above the horizontal. A short

Question

In a circus act, a 78 kg clown is shot from a cannon with an initial velocity of 19 m/s at some unknown angle above the horizontal. A short time later the clown lands in a net that is 4.4 m vertically above the clown’s initial position. Disregard air drag. What is the kinetic energy of the clown as he lands in the net?

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King 5 years 2021-08-12T21:06:46+00:00 1 Answers 14 views 0

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  1. haidang
    0
    2021-08-12T21:08:23+00:00
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    Answer:

    KE_2=10712.20\ J          

    Explanation:

    Given that

    Mass , m = 78 kg

    Initial velocity ,v= 19 m/s

    Vertical height  h= 4.4 m

    Now by using energy conservation

    Initial kinetic energy + Initial potential energy = Final kinetic energy +Final potential energy

    KE₁ + U₁ = KE₂ + U₂

    Therefore

    \dfrac{1}{2}mv^2+ 0 = KE_2+mgh

    Now by putting the values in the above equation

    \dfrac{1}{2}\times 78\times 19^2+ 0 = KE_2+78\times 9.81\times 4.4

    KE_2=\dfrac{1}{2}\times 78\times 19^2-78\times 9.81\times 4.4

    KE_2=10712.20\ J

    Therefore the final kinetic energy will be 10712.20 J.

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