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victus00 [196]
3 years ago
10

Two observers are 300 ft apart on opposite sides of a flagpole. The angles of elevation from the observers to the top of the pol

e are 16degrees and 20degrees. Find the height of the flagpole.
Physics
1 answer:
valentina_108 [34]3 years ago
3 0

Answer:

h = 48.077 ft

Explanation:

given,

distance between two observer = 300 ft

angle of elevation to top pole = 16° and 20°

height of the flagpole = ?

now,

Let h be the height of the flagpole

Let x be the distance of the pole

tan 16^0 = \dfrac{h}{x}

x =\dfrac{h}{tan 16^0}

now,

again applying

tan 20^0 = \dfrac{h}{300-x}

300-x=\dfrac{h}{tan 20^0}

300-3.49 h=2.75 h

6.24h = 300

h = 48.077 ft

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Daring Darless wishes to cross the Grand Canyon of the Snake River by being shot from a cannon. She wishes to be launched at 65
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Answer:

She must be launched with minimum speed of <u>57.67 m/s</u> to clear the 520 m gap.

Step-by-step explanation:

Given:

The angle of projection of the projectile is, \theta =65°

Range of the projectile is, R=520 m.

Acceleration due to gravity, g=9.8\ m/s^2

The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.

The range of projectile is given as:

R=\frac{v_{0}^2\sin2\theta}{g}

Plug in all the given values and solve for minimum speed, v_0.

520=\frac{v_{0}^2\sin(2(65))}{9.8}\\520\times 9.8=v_{0}^2\sin(130)\\5096=1.532v_{0}^2\\v_0^2=\frac{5096}{1.532}\\v_0^2=3326.371\\v_0=\sqrt{3326.371}=57.67\textrm{ m/s}

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A child jumps from a moving sled with a speed of 2.2 m/s and in the
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The initial velocity of the sled will be 7.34 m/sec. V is the initial velocity of the sled.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

(m₁)  mass of child = 38 kg

(u₁) is the initial velocity child = 2.2 m/s

(m₂) is the mass of sled = 68 kg

(u₂) is the initial velocity of sled = ?

(v) is the velocity after collision = 5.5 m/s

According to the law of conservation of momentum;

Momentum before collision =Momentum after collision

\rm  m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\(38 \times  2.2) + (68 \times V) = 5.5 \times (38+68) \\\\ 83.6 + 68V = 5.5 \times 106 \\\\  83.6 + 68V =583 \\\\ \rm 68\ V= 583 -83.6 \\\\ \rm 68 \ V= 499.4\\\\ V= 7.34 \ m/sec

Hence,the initial velocity of the sled will be 7.34 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

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