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pav-90 [236]
4 years ago
15

A projectile has an initial velocity of 110 m/s and a launch angle of 40o from the horizontal. The surrounding terrain is level,

and air friction is to be disregarded. What is the maximum elevation achieved by the projectile?
Physics
1 answer:
Nikolay [14]4 years ago
4 0

Answer:

255.09 m

Explanation:

Using

H = U²sin²θ/2g.................. Equation 1

Where H = maximum elevation, U = initial velocity, θ = Angle of projectile, g = acceleration due to gravity.

Given: U = 110 m/s, θ = 40°, g = 9.8 m/s².

Substitute into equation 1,

H = 110²sin²40/(2×9.8)

H = 12100(sin²40)/(19.6)

H = 12100(0.4132)/19.6

H = 255.09 m.

Hence the maximum elevation achieved by the projectile = 255.09 m

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Likurg_2 [28]

Answer:

k = 4422.35  KN/m

Explanation:

Given that

Frequency ,f= 29 Hz

m = 7.5 g

Natural frequency ω

ω = 2 π f

We also know that for spring mass system

ω ² m =k        

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So we can say that

( 2 π f)² =  m k

By putting the values

(2 x π x 29)² = 7.5 x 10⁻³  k

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k=4422.35 x  10³ N/m

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3 years ago
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Ilia_Sergeevich [38]

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3 years ago
A baseball is released at rest from the top of the Washington Monument. It hits the ground after falling for 6 s. What was the h
alukav5142 [94]

Answer:

Total height (s) = 176.4 m

Explanation:

Given:

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Ede4ka [16]

Answer:

The magnitude of the free-fall acceleration at the orbit of the Moon is 2.728\times 10^{-3}\,\frac{m}{s^{2}} (\frac{2.784}{10000}\cdot g, where g = 9.8\,\frac{m}{s^{2}}).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration (g), in meters per square second, is directly proportional to the mass of the Earth (M), in kilograms, and inversely proportional to the distance from the center of the Earth (r), in meters:

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G - Gravitational constant, in cubic meters per kilogram-square second.

M - Mass of the Earth, in kilograms.

r - Distance from the center of the Earth, in meters.

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