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Chapter 3: Problem 34

You toss an apple horizontally at \(8.7 \mathrm{m} / \mathrm{s}\) from a height of \(2.6 \mathrm{m}\) Simultaneously, you drop a peach from the same height. How long does each take to reach the ground?

Short Answer

Expert verified
Both the apple and the peach hit the ground at the same time which is approximately \(0.73\) seconds.

Step by step solution

01

Determine the Physics Concept

We are dealing with free-falling objects here. The only force they experience is gravity. Therefore we use the following equation of motion which applies to freely falling bodies: \( h = 0.5gt^2 \), where \( h \) is the height, \( g \) is the acceleration due to gravity (which is approximately \( 9.8 m/s^2 \) on Earth), and \( t \) is the time.
02

Insert Known Values

Insert the known values into the formula. So, for \( h = 2.6 m \) and \( g = 9.8 m/s^2 \), the equation becomes: \( 2.6 = 0.5 * 9.8 * t^2 \)
03

Solve for Time

Now solve for \( t \). First divide both sides of the equation by 0.5 * 9.8 (which is 4.9), to get \( t^2 = 2.6 / 4.9 \), then take square root to find \( t \).

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