Question

Question 2 A 50 kg skateboarder is planning a trick for an upcoming competition. She starts out riding her 7 kg skateboard with initial speed of 2.5 m/s. She and the skateboard then slide down a slope of height 3 meters. At the bottom of the slope, she steps from the first skateboard onto a second stationary 7 kg skateboard, leaving the first skateboard moving at half the speed it was before the switch. Finally, riding the 2nd skateboard, she goes up a quarter-pipe of height 2 meters. Neglecting friction, how high above the quarter-pipe could she go?

Answer 1

to find speed at bottom of slope, Phit kei = Pb + If ugh + 2 mug ² = 0 + 1 v2 v² = (2.5)² + (2x 3.8 X 3) u= 8.07 m/s Now PEP (to switch skateboard) (su + +)(8-7) – @x ) + (90+%) Using v= 7.57m18 Energy conservation, 12 mv² = mg (H+2) 7.57299) CH+2) TH = 0.924 m