need to find final height and kinetic energy

Question

a 3.8 kg shot put leaves the athletes hand with an initial velocity of 5.0 m/a straight up.

What is the initial kinetic energy of the shot put?

What is the final height of the shot put?

Answer 1

To solve this problem, we need to use the concepts of energy conservation and projectile motion.

1. Initial kinetic energy:

The initial kinetic energy of the shot put is given by:

K1 = (1/2)mv1^2

where m is the mass of the shot put, and v1 is the initial velocity.

Plugging in the values given in the problem, we get:

K1 = (1/2)(3.8 kg)(5.0 m/s)^2 = 47.5 J

Therefore, the initial kinetic energy of the shot put is 47.5 J.

1. Final height:

To find the final height of the shot put, we can use the fact that the total mechanical energy of the shot put (i.e., the sum of its kinetic and potential energies) is conserved throughout its motion. At the highest point of its trajectory, the shot put has zero kinetic energy, and its total energy is entirely in the form of potential energy.

The potential energy of the shot put at its maximum height is given by:

U = mgh

where m is the mass of the shot put, g is the acceleration due to gravity (9.8 m/s^2), and h is the maximum height.

At the highest point of its trajectory, the velocity of the shot put is zero, so the initial kinetic energy K1 is entirely converted into potential energy U.

Therefore:

K1 = U

(1/2)mv1^2 = mgh

Solving for h, we get:

h = (1/2)(v1^2/g) = (1/2)(5.0 m/s)^2 / 9.8 m/s^2 = 1.3 m

Therefore, the final height of the shot put is 1.3 m.