contestada

A 2 kg rocket is launched straight up into the air with a speed that allows it to reach a height of 100 meters, even though air resistance performs 800 J of work on the rocket. Determine the launch speed of the rocket. How high would the rocket travel if air resistance is ignored?

Respuesta :

Answer:

v = 52.5 m/s   and   h = 140.6 m

Explanation:

We use the energy work relationship

    W = ΔEm

W = - 800 J

The negative sign is because rubbing always opposes movement

Let's write it enegia in two points

Lower initial

     Em₀ = K = ½ m v²

Highest final

      [tex]Em_{f}[/tex] = U = mg h

     W =  [tex]Em_{f}[/tex] - Emo

     800 = ½ m v² - m g h

     v² = (800+ m gh) 2 / m

     v = √ (800 + 2 9.8 100) 2/2

     v = 52.5 ms

we  ignoring the existence of air the height is

      [tex]Em_{f}[/tex] = Em₀

     ½ m v² = m g h

     h = v² / 2g

     h = 52.5²/2 9.8

     h = 140.6 m

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