Cutnell, Physics, 11e Chapter 03, Problem 15 A skateboarder shoots off a ramp wi

Question

Cutnell, Physics, 11e Chapter 03, Problem 15 A skateboarder shoots off a ramp with a velocity of 7.8 m/s, directed at an the *y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontaly from the end of the ramp? angle of 63° above the horizontal. The end of the ramp is 1.3 m above the ground. Let the x axis be parallel to the ground the ground is the highest point that the skateboarder (a) Number Units (b) Number Units By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor Question Attempts: UnlimitedSAVE FOR LATER Earm Maximum Points available only ifyou answer this question correctly in four attempts or less. MacBook Pro

Explanation / Answer

The initial vertical velocity of the skateboarder=7.8 sin63
vertical acceleration = g = 9.8m/s2

Let h be the height skateboarder reach above the end of the ramp
   u =7.8 sin63
   v = 0
using equation
    v2 = u^2-2*g*h
     0 =(7.8 sin63)^2-2*9.8*h
=> h = 5.49 m

b)
     the highest point is 1.3 +5.49=6.79 m above the ground

       time taken to reach thehighest point=t
         v =u-g*t
          0 =7.8*sin63 - 9.8*t
          t =1.06 s

      initial horizontal velocity=7.8*cos63
      horizontal acceleration =0
distance traveled horizontally =7.8*cos63*1.06

= 3.75m

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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