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dolphi86 [110]
4 years ago
8

7) A kite 50 ft above the ground moves horizontally at a speed of 4 ft/sec. At what rate is the length of the string increasing

when 100 ft of string has been let out and the angle between the string and horizontal is decreasing at a rate of 0.3 rad/sec?
Physics
1 answer:
d1i1m1o1n [39]4 years ago
3 0

Imagine a right triangle where the legs represent the horizontal and vertical lengths of the string and the hypotenuse represents the length of the string.

Let us assign some values:

x = horizontal length in feet

50 = vertical length in feet

L = length of the string in feet

Because we are modeling these quantities with a right triangle, we can use the Pythagorean theorem to relate them with the following equation:

L² = x² + 50²

We want to find an equation for the change of L over time, so first differentiate both sides with respect to time t then solve for dL/dt:

2L(dL/dt) = 2x(dx/dt)

dL/dt = (x/L)(dx/dt)

First let's solve for x at the moment in time described in the problem using the Pythagorean theorem:

L² = x² + 50²

Given values:

L = 100ft

Plug in and solve for x:

100² = x² + 50²

x = 86.6ft

Now let's find dL/dt. Given values:

x = 86.6ft, L = 100ft, dx/dt = 4ft/sec

Plug in and solve for dL/dt:

dL/dt = (86.6/100)(4)

dL/dt = 3.46ft/sec

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v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}

v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}

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x=v_{x,i}t=v_{x,f}t

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s

The vertical component of the ball's velocity at time <em>t</em> is

v_{y,f}=v_{y,i}-gt

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}

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\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j

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and direction <em>θ</em> such that

\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height <em>y</em> at time <em>t</em> is

y=v_{y,i}t-\dfrac12gt^2

so that when it lands in the seats at <em>t</em> ≈ 6.38 s, it has a height of

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