Question

40 kg skateboarder enters a ramp moving horizontally with a speed of 9.6 m/s leaving the ramp vertically with a speed of 8.2 m/s. find the height of the ramp, assuming no energy loss to frictional forces. Solve from a energy perspecting and also draw a force diagram with all work shown.

Answer 1

initial kinetic energy of the skateboarder = K1 = 0.5*m*v1^2

initial potential energy of the skate boarder = U1 = 0

while leaving at height h

final kinetic energy of the skate boarder = K2 = 0.5*m*v2^2

final potential energy of the skate boarder = U2 = m*g*h

from energy conservation, total energy is conserved.

U1 + K1 = U2 + K2

0 + 0.5*m*v1^2 = m*g*h + 0.5*m*v2^2

v1 = 9.6

v2 = 8.2

m = 40

0.5*40*9.6^2 = 40*9.8*h + 0.5*40*8.2^2

h = 1.27 m <<-----answer