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Studentka2010 [4]

4 years ago

8

A ball thrown vertically upward is caught by the thrower after 4.00 sec. Find the initial velocity of the ball and the maximum h

eight it reaches.

1 answer:

My name is Ann [436]4 years ago

7 0

Answer:

Initial velocity = 39.2m/s

Maximum height is 78.4m

Explanation:

Given

$Time, t = 4s$

Solving (a): Initial Velocity

Using first law of motion:

$v = u + at$

Where

$v = final\ velocity = 0$

$u = iniital\ velocity = ??$

<em /> $a = acceleration = -g$ <em> [g represents acceleration due to gravity]</em>

$t = 4$

Substitute these value in the above formula:

$v = u + at$

$0 = u - g * 4$

$0 = u - 9.8 * 4$

Take g as 9.8m/s²

$0 = u - 39.2$

$u = 39.2m/s\\$

<em>Hence, initial velocity = 39.2m/s</em>

Solving (b): Maximum Height

This will be solved using second equation of motion

$s = ut + \frac{1}{2}at^2$

This becomes

$s = ut - \frac{1}{2}gt^2$

Substitute values for u, t and g

$s = 39.2 * 4 - \frac{1}{2} * 9.8 * 4^2$

$s = 156.8 - 78.4$

$s = 78.4$

<em>Hence, the maximum height is 78.4m</em>

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where,

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4 years ago

A hollow conductor is positively charged. Asmall uncharged metal ball is lowered by a silk thread through asmall opening in the

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Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

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$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

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Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

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If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

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$n = \frac{4.60517}{3.8918}=1.18$

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3 years ago