Question

The tomato is dropped. What is the velocity, v, of the tomato when it hits the ground? Assume 85.6 % of the work done in Part A is transferred to kinetic energy, E, by the time the tomato hits the ground.

Answer 1

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.

Answer 2

Answer:

$v = 4.1 \sqrt{h}$

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

$\frac{85.6}{100}\times m\times g\times h = \frac{1}{2}\times m\times v^{2}$

$v = 4.1 \sqrt{h}$