(a) How high above the ground is the highest point that theskateboarder reaches?

Question

(a)   How high above the ground is the highest point that theskateboarder reaches? (b)   When the skateboarder reaches the highest point, how far isthis point horizontally from the end of the ramp? A skateboarder shoots off a ramp with a velocity of 6.6 m/s,directed at an angle of 58° above the horizontal. The end ofthe ramp is 1.2 m above the ground. Let the x axis be parallel tothe ground, the +y direction be vertically upward, and take as theorigin the point on the ground directly below the top of the ramp. (a)   How high above the ground is the highest point that theskateboarder reaches? (b)   When the skateboarder reaches the highest point, how far isthis point horizontally from the end of the ramp?

Explanation / Answer

vertical acceleration = g = 9.8m/s2
let the skateboarder reach a height h above the end ofthe ramp    u =6.6*sin58    v = 0 using basic kinematic equation
    v2 = u2-2*g*h      0 =(6.6*sin58)2-2*9.8*h     ==> h = 1.598 m
b)
     the highest point is 1.2 +1.598=2.798 m above the ground
       time taken to reach thehighest point=t          v =u-g*t           0 =6.6*sin58 - 9.8*t           t =0.5711 s
      initial horizontalvelocity=6.6*cos58       horizontal acceleration =0     distance traveled in 0.478 s horizontally =6.6*cos58*0.5711 = 1.997 m

     the highest point is 1.997 m from theend of the ramp

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.