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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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3 years ago
A. A ball is thrown directly up with an initial speed of 4.00 m/s at y = 0. What is the maximum height h that it achieves, and w
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Answer:

A)     t = 0.40816 s , y = 0.916 m

Explanation:

A) For this problem we use the kinematic relations

           v = v₀ - g t

the highest point zero velocities (v = 0)

           t = (v₀-v) / g

           t = (4 - 0) / 9.8

           t = 0.40816 s

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          v² = v₀² - 2 g y

          y = vo2 / 2g

           y = 4 2 / (2 9.8)

          y = 0.916 m

To find the number of photos, we can use a direct proportions rule, if you take 30 photos in a second in 0.40816 s how many photos does it take

          # _photos1 = 0.40916 (30/1)

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          #_fotod = 0.40916 (120/1)

          #photos = 5.87 10³

 

B) The ball is released from a latura h how long it takes to reach the floor

           v² = v₀² + 2 g y

where the initial velocity is zero and the velocity with which the expert leaves is equal to the velocity with which v = 4 m / s leaves

            v² = 2gy

             v = √ (2 9.8 0.916)

             v = √ (2.1397 101)

             v = 4.6257 m / s

c) we ask us for the time for latura

             y = L / 2

             y = 0.916 / 2

             y = 0.458 m

now we can use the formula

             y = v₀ t - ½ g t²

           0.458 = 4.00 t - ½ 9.8 t²

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we solve second degree execution

           t = [0.8163 ±√ (0.8163² - 4 0.09346)] / 2

           t = [0.8163 ± 0.540] / 2

           t₁ = 0.678 m

           t₂ = 0.2763 m

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