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shtirl [24]
3 years ago
6

A projectile is fired in such a way that its horizontal range is equal to 14.5 times its maximum height. what is the angle of pr

ojection?
Physics
1 answer:
3241004551 [841]3 years ago
5 0

The maximum height is: hmax  = v0² sin² α / 2 g<span>
The horizontal range is: x = v0² sin 2 α / g</span>

So we are given that:

x = 14.5 hmax

v0² sin 2 α / g = 14.5 v0² sin² α / 2 g<span>
<span>7.25 sin² α = 2 sin α cos α    
<span>7.25 sin α = 2 cos α  
7.25 tan α = 2
<span>tan α = 0.27586
<span>α = tan^(-1) 0.27586
α = 15.42°<span>
<span>The angle of projection is 15.42</span></span><span>°</span></span></span></span></span></span>

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In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra
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A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

v=\sqrt{\frac{GM}{R+h}}

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B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

v=\sqrt{\frac{2GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula, we find:

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s

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