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A physics class conducting a research project on projectile motion constructs a device that can launch a cricket ball. The launching device is designed so that the ball can be launched at ground level with an initial velocity of 28 m/ s at an angle of 30° to the horizontal. Calculate the horizontal component of the velocity of the ball:a. initiallyb. after 1.0 sc. after 2.0 s

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Answer:

Explanation:

initial velocity, u = 28 m/s

Angle of projection, θ = 30°

The acceleration in horizontal direction is zero, so the horizontal component of velocity is constant.

Horizontal component of velocity, u cos θ = 28 x Cos 30 = 24.25 m/s

At t = 2 sec, the horizontal component of velocity = 24.25 m/s

At t = 3 sec, the horizontal component of velocity = 24.25 m/s

lhmarianateixeira

In this exercise we have to use the knowledge about oblique launch to calculate the components of velocity for each case, so we have that:

For all times we will find a velocity equal to 24.25 m/s

organizing the information given in the statement we have that:

  • initial velocity, u = 28 m/s
  • Angle of projection, θ = 30°
  • The acceleration in horizontal direction is zero
  • velocity is constant.

Knowing that the component can be written as:

[tex]u cos \theta = 28 * Cos 30 = 24.25 m/s[/tex]

So as the formula does not depend on time we have that for any value it will have a constant velocity.

See more about velocity at brainly.com/question/862972

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