The stone was thrown at an angle α = 30 to the horizon. At the top point of the trajectory

The stone was thrown at an angle α = 30 to the horizon. At the top point of the trajectory, the potential energy of the stone, measured from the surface of the earth, is equal to U = 15 J. Find the value of the kinetic energy K at this point.

Given:

a = 30 degrees – the angle at which the stone was thrown;

g = 10 m / s ^ 2 – acceleration of gravity;

U = 15 Joules – the potential energy of the stone at the highest point of the trajectory.

It is required to determine W (Joule) – the kinetic energy of the stone at the upper point of the trajectory.

The speed v, with which a stone was thrown at the surface of the earth, should be decomposed into 2 components: horizontal v * cos (a) and vertical v * sin (a). The vertical speed at the highest point of the trajectory is zero, so the kinetic energy at this point will be determined only by the horizontal component.

Let the stone have mass m. Then the height of its maximum point of the trajectory is:

h = U / (m * g).

The time it takes for the stone to fall to the ground is equal to:

t = (2 * h / g) ^ 0.5 = (2 * U / (m * g ^ 2) ^ 0.5 = 1 / g * (2 * U / m) ^ 0.5.

Then the speed that he will reach at the end of the path is equal to:

v1 = g * t = g * 1 / g * (2 * U / m) ^ 0.5 = (2 * U / m) ^ 0.5.

We found the vertical component of the speed. Then the speed itself will be equal to:

v = v1 / sin (a) = (2 * U / m) ^ 0.5 / 0.5 = 2 * (2 * U / m) ^ 0.5.

And its horizontal component will be equal to:

v * cos (a) = 2 * (2 * U / m) ^ 0.5 * 3 ^ 0.5 / 2 = (6 * U / m) ^ 0.5.

Then the kinetic energy at the maximum point will be equal to:

W = m * v ^ 2/2 = m * ((6 * U / m) ^ 0.5) ^ 2/2 = m * 6 * U / (2 * m) = 3 * U = 3 * 15 = 45 Joules.

Answer: the kinetic energy of the stone at the highest point will be 45 Joules.