A projectile fired upward from the Earth's surface will usually slow down, come momentarily to rest, and return to Earth. For a certain initial speed, however it will move upward forever, with its speed gradually decreasing to zero just as its distance from Earth approaches infinity. The initial speed for this case is called escape velocity. You can find the escape velocity v for the Earth or any other planet from which a projectile might be launched using conservation of energy. The projectile of mass m leaves the surface of the body of mass M and radius R with a kinetic energy Ki = mv²/2 and potential energy Ui = -GMm/R. When the projectile reaches infinity, it has zero potential energy and zero kinetic energy since we are seeking the minimum speed for escape. Thus Uf = 0 and Kf = 0. And from conservation of energy,
Ki + Ui = Kf + Uf
mv²/2 -GMm/R = 0
∴ v = √(2GM/R)
This is the expression for escape velocity.
Answer:
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Explanation:
Answer:
The speed at the end of the track = 27 m/s
The acceleration = 1.2 m/s²
Please find the Δx vs Δt, v vs Δt, a vs Δt
Explanation:
We have;
x = u·t + 1/2·a·t²
Where;
x = The distance = 300 m
u = The initial velocity = 0 m/s (Ball at rest)
t = The time taken = 22.4 s
Therefore;
300 = 0 + 1/2×a×22.4²
a = 2×300/22.4² = 1.19579 ≈ 1.2 m/s²
v = u + a×t
∴ v = 0 + 1.2 × 22.4 = 26.88 ≈ 27 m/s
Part of the table of values is as follows;
t, x, v
0, 0, 0
0.4, 0.095663, 0.478316
0.8, 0.382653, 0.956632
1.2, 0.860969, 1.434948
1.6, 1.530611, 1.913264
2, 2.39158, 2.39158
2.4, 3.443875, 2.869896
2.8, 4.687497, 3.348212
3.2, 6.122445, 3.826528
3.6, 7.748719, 4.304844
Answer:
Centre of mass of any body is a point where all mass of a body is supposed to be concentrated
it lies in geometrical centre....