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Alisiya [41]
2 years ago
12

A ball is thrown straight upward with an initial velocity of 18.5m/s at the edge of a cliff that is 25.0m below the initial poin

t. What is the velocity of the ball at its maximum height?
Physics
1 answer:
kkurt [141]2 years ago
8 0

The velocity of the ball at its maximum height is zero (0).

The given parameters;

  • <em>initial vertical velocity of the ball, </em>v_y_0<em> = 18.5 m/s</em>
  • <em>height of the cliff, h = 25</em>

<em />

In a projectile motion, as the object ascends upwards, its vertical velocity decreases and eventually becomes zero as the object reaches maximum height.

This <em>velocity</em> starts to increase again as the object descends downwards and finally becomes maximum before the object hits the ground.

Thus, we can conclude that the velocity of the ball at its maximum height is zero (0).

Learn more here:brainly.com/question/10693605

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Answer:

F=5.8\times 10^{8}\ N

F=35.57\times 10^{21}\ N

Explanation:

Given that

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A=\pi (6370\times 10^3)^2\ m^2

A=1.27\times 10^{14}\ m^2

As we know that pressure due to intensity given as

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F=4.6\times 10^{-6}\times 1.27\times 10^{14}\ N

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b)Gravitational force F

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M = mass\ of\ sun = 2\times 10^{30} kg\\m = mass\ of\ earth = 6\times 10^{24}kg

r =1.5\times 10^{11}\ m

G =6.67\times 10^{-11}Nm^2/kg^2

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F=6.67\times 10^{-11}\times \dfrac{2\times 10^{30}\times 6\times 10^{24}kg}{1.5\times 10^{11}}

F=35.57\times 10^{21}\ N

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