Question

A skateboarder starts up a 1.0-m-high, 30∘ ramp at a speed of 8.6 m/s . The skateboard wheels roll without friction. At the top, she leaves the ramp and sails through the air.

Part A

How far from the end of the ramp does the skateboarder touch down?

Express your answer to two significant figures and include the appropriate units.

Answer 1

The gravitational acceleration component along in the ramp (Aɢ) is:

  Aɢ = g • sin(30º)
  Aɢ = - 9.8 • sin(30º)
  Aɢ = -  4.9 m/sec²  ...  negative means in the down-ramp direction here.

The uphill distance traveled along the ramp (L) is:

  sin(30º)  =  h ⁄ L  =  (1 meter high) ⁄ L ————> L = 2 m

... so the ramp length (L) is only 2 m long

     (Vғ)² – (Vi)² = 2 • a • L

where;
   a = Aɢ = -  4.9 m/sec²
   Vғ = final velocity
   Vi = initial velocity = 8.6 m/sec

     (Vғ)² – (Vi)² = 2 • a • L
     (Vғ)² – (8.6)² = 2 • (-  4.9) • 2
         Vғ =9.67 m/sec  ...   velocity at the end of the ramp

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Therefore the skateboarder goes airborne at at launch velocity (Vʟ) of:

  Vʟ =9.67 m/sec at a 30º angle

where: Vʟx and Vʟy are the horizontal and vertical components of the vector Vʟ

  Hi = initial height = 1 m
  Hғ = final height = 0 (impact)

     Hғ  =  (½)  •  g  •  T ²   +   (Vʟy)  •  T   +   Hi

     Hғ  =  (½)  •  g  •  T ²   +   (Vʟ)  •  sin(30º)  •  T   +   Hi

     0  =  (½)  •  (- 9.8)  •  T ²   +   (½)  •  (9.67)  •  T   +   1

     0  =  (- 4.9) T ²  +  (1.457) T  +  1

       T = 0.624 sec  ... time of flight ... only positive time solution

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

 Dx = horizontal distance from the end of the ramp

     Dx = (Vʟx) • T  ...  time of flight = T

     Dx = (Vʟ) • cos(30º) • T

     Dx = (√ 3) • (Vʟ) • T ⁄ 2

     Dx = (√ 3) • 9.67) • T ⁄ 2

     Dx = (2.523) • T

     Dx = (2.523) • (0.624)

     Dx = 1.57 m