Answer:
(a)[tex]h=5.95m[/tex]
(b) h is the same
Explanation:
According to the law of conservation of energy:
[tex]E_i=E_f\\U_i+K_i=U_f+K_f[/tex]
The skier starts from rest, so [tex]K_i=0[/tex] and we choose the zero point of potential energy in the end of the ramp, so [tex]U_f=0[/tex]. We calculate the final speed, that is, the speed when the skier leaves the ramp:
[tex]mgH=\frac{mv^2}{2}\\v=\sqrt{2gH}\\v=\sqrt{2(9.8\frac{m}{s^2})(27m)}\\v=23\frac{m}{s}[/tex]
Finally, we calculate the maximum height h above the end of the ramp:
[tex]v_f^2=v_i^2-2gh\\[/tex]
The initial vertical speed is given by:
[tex]v_i=vsin\theta[/tex]
and the final speed is zero, solving for h:
[tex]h=\frac{v_i^2}{2g}\\h=\frac{((23\frac{m}{s})sin(28^\circ))^2}{2(9.8\frac{m}{s^2})}\\h=5.95m[/tex]
(b) We can observe that the height reached does not depend on the mass of the skier