Question

A student wishes to record a 7.5-kilogram watermelon colliding with the ground. Calculate how far the watermelon must fall freely from rest so it would be traveling at 29 meters per second the instant it hits the ground. [Show all work, including the equation and substitution with units.]

Answer 1

Answer:

42.86m

Explanation:

The first thing we should keep in mind is that the watermelon moves with uniform acceleracion equal to gravity (9.81m / s ^ 2)

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

$\frac {Vf^{2}-Vo^2}{2.g} =Y$

where

Vf=29m/s= final speed

Vo= initial speed=0m/S

g=gravity=9.81m/s^2

Y= distance traveled(m)

solving

$\frac {Vf^{2}-Vo^2}{2.g} =Y\\\frac {(29m/s)^{2}-(0m/s)^2}{2(9.8m/s^2)} =Y\\Y=42.86m$

the distance traveled by watermelon is 42.86m