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3 years ago

The velocity at the midway point of a ball able to reach a height y when thrown with velocity vi at the origin is:

1 answer:

Nutka1998 [239]3 years ago

7 0

Assuming the ball follows classical 2D projectile motion (moves in a parabola) and that the height y = the maximum height the ball goes in the y direction (because this would be its midpoint), then the velocity at height y is equal to the initial x component of velocity. At the midpoint, the y component is zero, so the velocity only depends on the x component. Projectiles move at constant speed in the x direction, so X = Xo. As long as you know actual values for Vi and either the initial angle or one initial component, then you can solve for Xo using trigonometry. Xo is thus the velocity of the ball once it has reached its maximum height.

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Solution :

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$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

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Acceleration due to gravity at 2000 km depths is :

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$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

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Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

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Acceleration due to gravity at 4000 km depths is :

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Answer:

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