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sergij07 [2.7K]

3 years ago

6

An arcade ball is thrown with an initial speed of 7.0 m/s and follows the trajectory shown. The ball enter the basket .95 second

s after it is launched. What are the distances x and y?

1 answer:

dmitriy555 [2]3 years ago

7 0

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

$x = 4.70 \ m$ , $y = 0.2803 \ m$

Explanation:

From the question we are told that

The initial velocity of the arcade ball is $u = 7.0 \ m/s$

The time taken by the ball to enter the basket is $t = 0.95 \ s$

Generally the x -component of the initial velocity is

$u_x = 7 * cos (45)$

=> $u_x = 4.95 \ m/s$

Generally the y -component of the initial velocity is

$u_x = 7 * sin(45)$

=> $u_y = 4.95 \ m/s$

Generally the distance x is mathematically represented as

$x = u_x * t$

=> $x = 4.95 * 0.95$

=> $x = 4.70 \ m$

Generally from kinematic equation the distance y is mathematically represented as

$y = u_y * t - \frac{1}{2} * g * t^2$

=> $y = 4.95 * 0.95 - \frac{1}{2} * 9.8 * 0.95^2$

=> $y = 0.2803 \ m$

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Answer:

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Explanation:

As per the law of uncertainty we know that

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now we know that

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also we have

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now we will have

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Answer:

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Explanation:

The illustration of the problem is shown the attached image.

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$L^2 = BC^2 + (24 + AE)^2$ ........1

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$\frac{BC}{8} = \frac{AE + 24}{AE}$

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Have a nice day!

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