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2 years ago

6

From the top of a vertical pole the angle of depression to Ian standing at the foot of the pole is 43°. Liam is on the other sid

e of the pole, and the angle of depression from the top of the pole to Liam is 52°. The boys are standing 58 m apart. Find the height of the pole, to the nearest meter

2 answers:

kifflom [539]2 years ago

6 0

The height of the pole based on the angle of depression given is 62.2m.

<h3>How to calculate the height?</h3>

From the information given, Liam is on the other side of the pole, and the angle of depression from the top of the pole to Liam is 52° and the boys are standing 58 m apart.

Let the height = h

It should be noted that:

tan = opposite/adjacent

tan 43° = 58/h

tan 43° × h = 58

0.9325 × h = 58

0.9325h = 58

h = 58/0.9325

h = 62.2m

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Tanya [424]2 years ago

3 0

The height of the pole is 26.94m

<h3>What is angle of depression?</h3>

Angle of depression is the angle between the line of sight and the horizontal.

Analysis:

let the distance from pole to the other guy be x

tan 38 = h/x

Also

tan 47 = h/58- x

(58 - x ) tan47 = xtan 38

(58 - x)(1.072) = 0.78x

62.17 - 1.072x = 0.78x

62.17 = 1.072x+0.78x

1.8x = 62.17

x = 34.54

h = xtan38

h = 34.54(0.78) = 26.94m

In conclusion, the height of pole is 26.94m

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Answer:

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Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

$\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j$

The particular expression for the cannonball is:

$\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$

b) The components of the position of the cannonball before hitting the ground is:

$x = (90\cdot \cos \theta)\cdot t$

$0 = 90\cdot \sin \theta - 6\cdot t$

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

$0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)$

$0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x$

$0 = 4050\cdot \sin 2\theta - 6\cdot x$

$6\cdot x = 4050\cdot \sin 2\theta$

$x = 675\cdot \sin 2\theta$

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

$\frac{dx}{d\theta} = 1350\cdot \cos 2\theta$

$1350\cdot \cos 2\theta = 0$

$\cos 2\theta = 0$

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

$\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta$

$\frac{d^{2}x}{d\theta^{2}} = -2700$

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

$y = 45\cdot t - 6\cdot t^{2}$

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

$\frac{dy}{dt} = 45 - 12\cdot t$

$45-12\cdot t = 0$

$t = \frac{45}{12}$

$t = 3.75\,s$

Now, the second derivative is used to check if such solution leads to a maximum:

$\frac{d^{2}y}{dt^{2}} = -12$

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

$y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}$

$y_{max} = 84.375\,m$

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