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A bowling ball is traveling at 3 m/s. It starts to roll up a ramp. How high above the ground will the ball be when it stops rolling? Neglect friction and assume the ramp is plenty long enough to do this. (For this problem your initial kinetic energy can be set equal to your final potential energy. 1/2mv2 = mgh).

Respuesta :

Answer:

h= 0.45 m

Explanation:

PE= 1/2 mv^2             KE= mgh

v= 3m/s

vf= 0 m/s

h=?

PE= 1/2(1kg)(3m/s)^2

PE= 4.5 J

4.5 J/ 1kg(9.8 m/s^2)

h=0.45 m

asuwafohjames

The height of the ball above the ground when it stops is 0.46 m.

The initial kinetic energy of the ball is equal to the final potential energy of the ball.

⇒ Formula:

  • mv²/2 = mgh
  • v² = 2gh............... Equation 1

⇒ Where:

  • v = velocity of the ball
  • h = height of the ball above the ground
  • g = acceleration due to gravity.

⇒ make h the subject of the equation

  • h = v²/2g.............. Equation 2

From the question,

⇒ Given:

  • v = 3 m/s
  • g = 9.8 m/s²

⇒ Substitute these values into equation 2

  • h = 3²/(2×9.8)
  • h = 9/19.6
  • h = 0.46 m

Hence, the height of the ball above the ground when it stops is 0.46 m.

Learn more about height here: https://brainly.com/question/12446886

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