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A 10.0kg ball is falls from a window 7.00m above the ground. What will be the velocity of the ball as it strikes the ground?

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Final answer:

The velocity of a 10.0 kg ball as it strikes the ground after falling from a 7.00 m height is approximately 11.72 m/s towards the ground, calculated using the free fall motion equation.

Step-by-step explanation:

The question asks for the velocity of a 10.0 kg ball as it strikes the ground after falling from a 7.00 m height. To find this, one can use the equation of motion under the influence of gravity, which is free fall. The equation v^2 = u^2 + 2gh can be used where v is the final velocity, u is the initial velocity (which is 0 for a free fall), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (7.00 m).

Plugging in the values: v^2 = 0 + 2 * 9.8 m/s^2 * 7.00 m gives us v^2 = 137.2 m^2/s^2. Therefore, v = √137.2 m^2/s^2, which is approximately 11.72 m/s towards the ground. This is the velocity of the ball when it strikes the ground.

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