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A ball is thrown toward a cliff of height h with a speed of 32 m/s and an angle of 60∘ above horizontal. It lands on the edge of the cliff 3.1 s later

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onyebuchinnaji

The height of the cliff is determined as 33.44 m.

What is height of a projectile?

The maximum height of a projectile above its launch position, is the the vertical distance of the projectile and it depends only on the vertical component of the initial velocity.

Height of the cliff

The height of the cliff is calculated as follows;

h(y) = v(y)t - ¹/₂gt²

where;

  • h is height of the cliff
  • v(y) is the initial vertical velocity of the ball
  • t is time of motion of the ball

h(y) = (32 x sin 60)(3.1) - (0.5)(9.8)(3.1²)

h(y) = 80.54  -  47.1

h(y) = 33.44 m

Thus, the height of the cliff is determined as 33.44 m.

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

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