Question

the 70.0 kg skateboarder in the drawing starts down the left side of the ramp with an initial speed of 4.50 m/s. neglect nonconservative forces, such as friction and air resistance. what is the height h of the skateboarder, when he has reached his maximum height on the right side of the ramp? hint: the speed of an object at its maximum height is zero. 5.21 m 3.09 m 1.03 m 4.67 m

Explanation:

Step1: Apply the conservation of mechanical energy

The initial mechanical energy \(E_1\) is the sum of kinetic energy \(K_1=\frac{1}{2}mv_1^2\) and potential energy \(U_1 = mgh_1\). Let the initial height \(h_1 = 0\) (reference level), so \(E_1=\frac{1}{2}mv_1^2\). At the maximum height \(h\) (final state), the speed \(v_2 = 0\), and the mechanical energy \(E_2=mgh\). By the conservation of mechanical energy \(E_1 = E_2\), we have \(\frac{1}{2}mv_1^2=mgh\).

Step2: Solve for \(h\)

Cancel out the mass \(m\) from both sides of the equation \(\frac{1}{2}mv_1^2=mgh\). We get \(h=\frac{v_1^2}{2g}\). Given \(v_1 = 4.50\space m/s\) and \(g = 9.8\space m/s^2\), substitute the values: \(h=\frac{(4.50)^2}{2\times9.8}=\frac{20.25}{19.6}\approx1.03\space m\)

Answer:

\(1.03\space m\)