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A tennis player at the Australian Open tosses a tennis ball straight up and then catches it after 1.63 s at the same height as the point of release. What is the acceleration of the ball while it is in flight?

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Final Answer:

The acceleration of the tennis ball while it is in flight is approximately -9.81 m/s².

Step-by-step explanation:

Define the given information:

Time of flight (t) = 1.63 s

Initial and final height are the same (h1 = h2)

Identify the relevant formula:

The vertical motion of the tennis ball is governed by the following equation:

h = h_0 + v_0 * t + (1/2) * a * t^2

where:

h is the height at any time t

h_0 is the initial height

v_0 is the initial velocity

a is the acceleration

t is the time

Analyze the problem:

Since the initial and final heights are the same (h1 = h2), the displacement (Δh) is zero. Therefore:

Δh = h2 - h1 = 0

Solve for the acceleration:

Substitute the known values into the equation and solve for a:

0 = 0 + v_0 * 1.63 + (1/2) * a * 1.63^2

2.6569a + 1.63v_0 = 0

We still don't have enough information to solve for a uniquely. However, we know that the acceleration due to gravity acts downwards and is approximately -9.81 m/s². Assuming no air resistance, we can use this value for a as an approximation.

Calculate the final value:

a ≈ -9.81 m/s²

Therefore, the acceleration of the tennis ball while it is in flight is approximately -9.81 m/s². This negative sign indicates that the acceleration is directed downwards.

Note: The initial velocity (v_0) is not required to solve for the acceleration in this case. However, if v_0 was known, we could solve for it as well.

User Becca Petrin
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