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anastassius [24]
3 years ago
11

1. A rocket hits the ground at an angle of 60° from the horizontal at a speed of 300 m/s.

Physics
1 answer:
alekssr [168]3 years ago
5 0

Answer:

(a). The horizontal and vertical components are v\cos\theta and v\sin\theta

(b). The horizontal and vertical components of the rocket's impact velocity is 150 m/s and 259.8 m/s.

Explanation:

Given that,

Angle = 60°

Speed = 300 m/s

(a). We need to draw the vector representing the rocket's impact and its components

Using given data

The components of rocket's impact

The horizontal component is

u_{x}=u\cos\theta

The vertical component is

v_{v}=v\sin\theta

(b). We need to calculate the horizontal and vertical components of the rocket's impact velocity

Using horizontal and vertical components

The horizontal component is

v_{x}=v\cos\theta

Put the value into the formula

v_{x}=300\cos60

v_{x}=150\ m/s

The vertical component is

v_{x}=v\sin\theta

Put the value into the formula

v_{x}=300\sin60

v_{x}=259.8\ m/s

Hence, (a). The horizontal and vertical components are v\cos\theta and v\sin\theta

(b). The horizontal and vertical components of the rocket's impact velocity is 150 m/s and 259.8 m/s.

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The 49-g arrow is launched so that it hits and embeds in a 1.45 kg block. The block hangs from strings. After the arrow joins th
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Answer:

the initial speed of the arrow before joining the block is 89.85 m/s

Explanation:

Given;

mass of the arrow, m₁ = 49 g = 0.049 kg

mass of block, m₂ = 1.45 kg

height reached by the arrow and the block, h = 0.44 m

The gravitational potential energy of the block and arrow system;

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P.E = (1.45 + 0.049) x 9.8 x 0.44

P.E = 6.464 J

The final velocity of the system after collision is calculated as;

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6.464 = ¹/₂(1.45 + 0.049)v²

6.464 = 0.7495v²

v² = 6.464 / 0.7495

v² = 8.6244

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Apply principle of conservation of linear momentum to determine the initial speed of the arrow;

P_{initial} = P_{final}\\\\mv_{arrow} + mv_{block} = (m_1 + m_2)V\\\\0.049(v) + 1.45(0) = (0.049 + 1.45)2.937\\\\0.049v = 4.4026\\\\v = \frac{4.4026}{0.049} \\\\v = 89.85 \ m/s

Therefore, the initial speed of the arrow before joining the block is 89.85 m/s

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