Answer:
[tex]v = 4.1 \sqrt{h}[/tex]
Explanation:
Let the mass of tomato is m and the height from which it falls is h.
Let the tomato its the ground with velocity v.
The potential energy of the tomato at height h
U = m x g x h
The kinetic energy of tomato as it hits the ground
K = 1/2 mv^2
According to the question,
85.6 % of Potential energy = Kinetic energy
[tex]\frac{85.6}{100}\times m\times g\times h = \frac{1}{2}\times m\times v^{2}[/tex]
[tex]v = 4.1 \sqrt{h}[/tex]
Answer:
Ok, we know that you drop a tomato of mass M from a height h, because you drop it, it has not initial velocity.
Now, the kinetic energy is:
K = (M/2)*v^2
because at the begginig there is not velocity, the kinetic energy is zero.
the potential energy is:
U = M*g*h
where g = 9.8m/s^2
We know that when the tomato hits the ground, the 85.6% of these potential energy is converted in kinetic; so we have:
K = 0.856*M*g*h = (1/2)*M*v^2
M can be canceled in both sides:
v^2 = 2*0.856*g*h
v = √(1.712*g*h)
is the velocity of the tomato when it hits the ground.
