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sashaice [31]
2 years ago
5

A hobby rocket reaches a height of 72.3 m and lands 111 m from the launch point with no air resistance. What was the angle of la

unch
Physics
1 answer:
Deffense [45]2 years ago
7 0

Answer:

The angle of launch is 52.49 Degree.

Explanation:

The Range R and Height H of a thrown object is calculated using the formula,

R=V₀² sin(2φ)/g

H=V₀²sin²(φ)/g

From these equations it can be written,

V₀²=R g/ sin(2φ)

V₀²=H g/ sin²(φ)

These values are equal so it can be written by equating these equations,

R g/sin(2φ)=H g/sin²(φ)

tan(φ)= 2H/R

Given H=72.3 m and R=111 m, the angle of launch is,

tan(φ)= 2*72.3/111

φ= 52.49 Degree.

Check out other solutions,

brainly.com/question/1495042

#SPJ10

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T^{2} = r^{3}

T = \sqrt{(133370 Km)^{3}}

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T = 4.870x10^{7} Km

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1 AU is defined as the distance between the earth and the sun.

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That can be expressed in units of days

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<em>Circular velocity for the particle in the </em><em>Encke Division</em><em>:</em>

v = \frac{2 \pi r}{T}

v = \frac{2 \pi (133370 Km)}{(118.60 days)}

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

118.60 days .\frac{86400 s}{1 day} ⇒ 10247040 s

133370 Km .\frac{1000 m}{1 Km} ⇒ 133370000 m

v = \frac{2 \pi (133370000 m)}{(10247040 s)}

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Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

T^{2} = r^{3}

T = \sqrt{(69000 Km)^{3}}

T = \sqrt{(3.285x10^{14} Km)}

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v = 114.483 m/s

 

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