Answer:
The answer to your question is: vo = 25 m/s
Explanation:
data
a = -7.5 m/s²
d = 42 m
vf = 0 m/s
vo = ?
Formula
vf² = vo² - 2ad
Substitution
0² = vo² - 2(7.5)(42)
We clear vo from the equation
vo² = 2(7.5)(42)
vo² = 630 simplifying
vo = 25 m/s result
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Answer:
Explanation:
Given that height of the projectile as a function of time is
here we know that
h = 147 ft
so from above equation
now by solving above quadratic equation we know that
elasticity stretches and can also return to it's normal size ..
If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:
Fg=Fcp
m*g=m*(v²/R),
where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite.
Masses cancel out and we have:
G*(M/r²)=v²/R, R=r so:
G*(M/r)=v²
r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,
r=G*(M/ω²r²),
r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:
r³=G*(M/(4π²/T²)), and finally we take the third root to get r:
r=∛{(G*M*T²)/(4π²)}=4.226*10^7 m= 42 260 km which is the height of a geostationary satellite.
