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 will hit the ground at the same time; approximately 0.73 seconds after being dropped/thrown.

Step by step solution

01

Understand the principle of free fall

Gravity, which is the force that attracts objects towards each other, is primarily what causes objects to fall. Any object that falls under gravity falls at the same rate, regardless of the horizontal velocity. So whether an object is dropped straight from a height or thrown horizontally, it will hit the ground at the same time assuming it starts from the same height and there is no air resistance.
02

Establish the relevant equations

The equation that links the distance travelled under gravity (the height in this case), the gravity and the time is: \(d = \frac{1}{2}gt^2\), where \(d\) is the distance (or height), \(g\) is the acceleration due to gravity and \(t\) is the time. We’re solving for \(t\), so after re-arranging, \(t = \sqrt{\frac{2d}{g}}\). In this case, \(d = 2.6 \mathrm{m}\) and \(g\) is roughly \(9.8 \mathrm{m/s^2}\) (depending on your location on earth).
03

Calculate the time it takes for the objects to hit the ground

Using the rearranged equation from Step 2, calculate the time it takes for the apple and the peach to hit the ground. \(t = \sqrt{\frac{2*2.6}{9.8}}\)

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