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Chapter 2: Problem 75

A stuntman sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is \(10.0 \mathrm{~m} / \mathrm{s}\), and the man is initially \(3.00 \mathrm{~m}\) above the level of the saddle. (a) What must be the horizontal distance between the saddle and the limb when the man makes his move? (b) How long is he in the air?

Short Answer

Expert verified
(a) The horizontal distance between the saddle and the limb when the man makes his move should be the distance calculated in Step 2. (b) The time he is in the air will be the time calculated in Step 1.

Step by step solution

01

Determine the time to fall

The time it takes for the stuntman to fall can be calculated using the equation of motion \( h = \frac{1}{2}gt^2 \), where \( h = 3.00 \, m \) is the height of the tree limb from the saddle and \( g = 9.8 \, m/s^2 \) is the acceleration due to gravity. We solve for \( t \), getting \( t = \sqrt{\frac{2h}{g}} \).
02

Calculate the horizontal distance

The horizontal distance between the saddle and the limb when the stuntman makes his move is given by \( d = vt \), where \( v = 10.0 \, m/s \) is the speed of the horse and \( t \) is the time calculated in Step 1. Substituting the values we get the distance.
03

Answer (a)

The horizontal distance from Part (a) can be concluded as the distance calculated in Step 2.
04

Answer (b)

The time he is in the air, from Part (b), will be the time calculated in Step 1.

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