At the base of a frictionless icy hill that rises at 23.0m above the horizontal, a toboggan has a speed of 12.0m/s toward the hill. How high vertically above the base will it go before stopping?

thats 23 degrees

To solve this problem, we can use the conservation of mechanical energy principle. The toboggan starts with kinetic energy at the base of the hill and rises, converting its kinetic energy into potential energy. At the highest point, the toboggan comes to rest, meaning it has zero kinetic energy and maximum potential energy.

To find the potential energy of the toboggan at the highest point, we need to use the conservation of mechanical energy equation, which states:

Initial kinetic energy + Initial potential energy = Final kinetic energy + Final potential energy

The initial kinetic energy is given as half the mass of the toboggan multiplied by its initial velocity squared, which can be written as:

K_initial = (1/2) * m * v_initial^2

v_initial is the initial velocity of the toboggan, given as 12.0 m/s.

The final kinetic energy is 0 because the toboggan comes to rest at the highest point.

Therefore, the equation becomes:

K_initial + U_initial = 0 + U_final

Where U_initial is the initial potential energy at the base of the hill and U_final is the final potential energy at the highest point.

The potential energy is given by the equation:

U = m * g * h

Where m is the mass of the toboggan, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the vertical height above the base.

Now we can rearrange the equation and solve for h:

[(1/2) * m * v_initial^2] + [m * g * 0] = 0 + [m * g * h]

Since the mass of the toboggan cancels out from both sides, we can divide the equation by m * g:

(1/2) * v_initial^2 = h

Plugging in the values given in the question:

(1/2) * (12.0 m/s)^2 = h

Simplifying the equation:

(1/2) * 144.0 m^2/s^2 = h

72.0 m^2/s^2 = h

Taking the square root of both sides to solve for h:

√(72.0 m^2/s^2) = √h

8.49 m ≈ h

Therefore, the toboggan will go approximately 8.49 m vertically above the base before stopping.

1.67m