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Licemer1 [7]
3 years ago
6

A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40

° above the horizontal. How far above or below its original level will the ball strike the opposite wall?
Physics
1 answer:
Vinvika [58]3 years ago
5 0

Answer:

Ball hit the tall building 50 m away below 10.20 m its original level

Explanation:

Horizontal speed = 20 cos40 = 15.32 m/s

Horizontal displacement = 50 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    50 = 15.32 t + 0.5 x 0 x t²

     t = 3.26 s

Now we need to find how much vertical distance ball travels in 3.26 s.

Initial vertical speed  = 20 sin40 = 12.86 m/s

Time = 3.26 s

Vertical acceleration = -9.81 m/s²

Substituting in s = ut + 0.5at²

    s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²

    s = -10.20 m

So ball hit the tall building 50 m away below 10.20 m its original level

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Answer:

The horizontal component of the truck's velocity is: 23.70 m/s

The vertical component of the truck's velocity is: 3.13 m/s

Explanation:

You have to apply trigonometric identities for a right triangle (because the ramp can be seen as a right triangle where the speed is the hypotenuse), in order to obtain the components of the velocity vector.

The identities are:

Cosα= \frac{CA}{H}

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Where H is the hypotenuse, α is the angle, CA is the adjacent cathetus and CO is the opposite cathetus

The horizontal component of the truck's velocity is:

Let Vx represent it.

In this case, CA=Vx, H=24 and α=7.5 degrees

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Let Vy represent it.

In this case, CO=Vy, H=24 and α=7.5 degrees

Vy=(24)Sen(7.5)

Vy=3.13 m/s

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3 years ago
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VikaD [51]

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Explanation:

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Aleksandr [31]

Answer:

Explanation:

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