<u>Answer:</u>
The height of ramp = 124.694 m
<u>Explanation:</u>
Using second equation of motion,

From the question,
u = 31 m/s; s = 156.3 m, a=0
substituting values

t = 
= 5.042 s
Similary, for the case of landing
t = 5.042 s; initial velocity, u =0
acceleration = acceleration due to gravity, g = 9.81 
Substituting in 

h = 124.694 m
So height of ramp = 124.694 m
Magna Carta. I’m pretty sure it’s the correct answer.
Joules is a unit for work which may decomposed into N.m. Work is a quantity which is a product of force (in this case, the woman's weight) and the distance she has traveled.
W = F x d ; d = W / F
Substituting the given,
d = (3.5 x 10^4 J) / (55 kg x 9.8 m/s²) = 64.94 m
Thus, the woman can climb up to 64.94 meters.
Answer:
a)5.88J
b)-5.88J
c)0.78m
d)0.24m
Explanation:
a) W by the block on spring is given by
W=
kx² =
(530)(0.149)² = 5.88 J
b) Workdone by the spring = - Workdone by the block = -5.88J
c) Taking x = 0 at the contact point we have U top = U bottom
So, mg
=
kx² - mgx
And,
= (
kx² - mgx
)/(mg) =
]/(0.645x9.8)
= 0.78m
d) Now, if the initial initial height of block is 3
= 3 x 0.78 = 2.34m
then,
kx² - mgx - mg
=0
(530)x² - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0
265x² - 6.321x - 14.8 = 0
a=265
b=-6.321
c=-14.8
By using quadratic eq. formula, we'll have the roots
x= 0.24 or x=-0.225
Considering only positive root:
x= 0.24m (maximum compression of the spring)
The answer is a 20 or 140 u choose because u subtract or add these numbers and sorry I can’t speak sri lanka or Indian but I know how to read them