Problem 6.86 Part A If friction and air resistance can be neglected, determine h
ID: 1659657 • Letter: P
Question
Problem 6.86 Part A If friction and air resistance can be neglected, determine her speed B when she reaches the horizontal end of the ramp at B A 55-kg skier starts from rest at the top of a ski jump point A in the figure (Figure 1), and travels down the ramp. Take that 36 m Express your answer to two significant figures and include the appropriate units BValue Units Figure 1 of 1 Submit My Answers Give Up Part B Determine the distance s to where she strikes the ground at C Express your answer to two significant figures and include the appropriate units 4 m .S alueUnits Submit My Answers Give UpExplanation / Answer
Let:
H = vertical drop from A to B = 36m - 4.4m = 31.6m
h = drop from B to the top of the landing slope = 4.4 m
= angle of the landing slope = 30º
=> sin = 1/2
cos = 3 /2
cos2() = 3/4
sin/cos2() = 2/3
g = acceleration of gravity = 9.81 m/s2
To Be Found: -
VB = speed of skier at B
S = slant distance of jump
and place:
origin of coordinates at B
x-axis horizontally
z-axis vertically
Segment I: A to B; conversion of potential to kinetic energy:
VB2 = 2gH
VB = (2gH) = 24.88m/s
Segment II: B to C; free fall
x = VB t
z = -(g/2)t2
= -(g/[2VB2]) x2 = -x2/(4H)
Equating vertical components:
h + s sin = x2/(4H)
x2 = 4H(h + s sin)
Equating horizontal components:
x = s cos = 2[H(h + s sin)]
x2 = s2 cos2() = 4H(h + s sin)
Solve this as a quadratic in s:
s2 - 4H(sin/cos2())s - 4Hh/cos2()
= s2 - (8H/3)s - 16Hh/3 = 0
s = 4H/3 + [(4H/3)2 + 16Hh/3] = (4H/3)(1 + [1 + 3h/H])
= 126.4/3 m (1 + (1 + 0.418))
= 42.13m ( 1+1.19)
= 92.26 m zx