As you found out while solving Part C, the only force acting on Tarzan that does work on him is gravity. An important property of gravity is that the work done by this force on a body moving along a curved path is the same as if the body were displaced vertically. Therefore, you can evaluate your previous work using the kinematics equations for free fall. What vertical distance Δy does a free-falling particle travel from the moment it starts to the moment it reaches a speed of 7.9 m/s if it starts from rest? Work out your solution using one of the equations for vertical motion with constant acceleration, specifically, v2f=v2i+2aΔywhere vi and vf are, respectively, the particle's initial and final speed, and a is the particle's acceleration.

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Physics

19 September, 22:08

As you found out while solving Part C, the only force acting on Tarzan that does work on him is gravity. An important property of gravity is that the work done by this force on a body moving along a curved path is the same as if the body were displaced vertically. Therefore, you can evaluate your previous work using the kinematics equations for free fall. What vertical distance Δy does a free-falling particle travel from the moment it starts to the moment it reaches a speed of 7.9 m/s if it starts from rest? Work out your solution using one of the equations for vertical motion with constant acceleration, specifically, v2f=v2i+2aΔywhere vi and vf are, respectively, the particle's initial and final speed, and a is the particle's acceleration.

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Answers (1)

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Colby Hinton · Answer 1

The vertical distance traveled is 3.18 m.

Explanation:

In order to find the vertical distance traveled by a body in free fall motion, starting from rest to a velocity of 7.9 m/s. For this purpose, we use Newton's third equation of motion,

2as = Vf² - Vi²

where,

a = g = 9.8 m/s²

s = vertical distance = Δy = ?

Vf = final velocity = 7.9 m/s

Vi = initial velocity = 0 m/s

Therefore,

2 (9.8 m/s²) Δy = (7.9 m/s) ² - (0 m/s) ²

Δy = (7.9 m/s) ² / (2) (9.8 m/s²)

Δy = 3.18 m