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dolphi86 [110]
3 years ago
5

Which ball has the greater acceleration at the instant of release?

Physics
1 answer:
yuradex [85]3 years ago
6 0

They're equal.

Ignoring air resistance, any object you're not holding onto has the same, constant acceleration ... 9.8 m/s^2 down ... from the moment it leaves your hand until the moment it hits the ground.

It doesn't matter how heavy the object is, and it doesn't matter whether you throw it up, down, or sideways, or just let go and let it drop.

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Can you come up with a mathematical relationship, based on your data that shows the relationship between distance from the charg
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Explanation:

This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.

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A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".

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4 years ago
A basketball player throws the ball at a 47 angle above the horizontal to a hoop which is located a horizontal distance L = 5.0
FromTheMoon [43]

Answer:

v_0 =1.71

Explanation:

the parabolic movment is described by the following equation:

y = tan(a)x-\frac{1}{2v_0^2(cos(a))^2}gx^2

where y is the height of the ball, a is the angle of launch, v_0 the initial velocity, g the gravity and x is the horizontal distance of the ball.

So, if we want that the ball reach the hood, we will replace values on the equation as:

0.8 = tan(47)(5)-\frac{1}{2v_0^2(cos(47))^2}(9.8)(5)^2

Finally, solving for v_0, we get:

v_0=\sqrt{\frac{-9.8(5)^2}{(0.8-tan(47)(5))2cos^2(47)}}

v_0 =1.71

4 0
4 years ago
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