A person standing on a vertical cliff a height h above the lake wants to jump into the lake but notices a rock just at the surface level with its furthest edge a distance s from the shore .The person realizes that with a running start it will be possible to just clear the rock, so the person steps back from the edge a distance d and starting from the rest ,runs at an acceleration that varies in time according to ax=b1t and then leaves the cliff horizontally. The person just clears the rock.

Question

A person standing on a vertical cliff a height h above the lake wants to jump into the lake but notices a rock just at the surface level with its furthest edge a distance s from the shore .The person realizes that with a running start it will be possible to just clear the rock, so the person steps back from the edge a distance d and starting from the rest ,runs at an acceleration that varies in time according to ax=b1t and then leaves the cliff horizontally. The person just clears the rock.

Find s in terms of the given quantities d ,b1,h,and acceleration due to gravity g.You may neglect all air resistance.

Answer 1

time to fall h meters

h=1/2 g t^2 or
t= sqrt(2h/g)

now, time to traverse that distance horizontally in time t.

s=v*t solve for v

but v^2=2*d*a1
or
s= sqrt(2d*a1)*sqrt(2h/g)
s=sqrt(2d*a1*2h/g)

check my thinking

Answer 2

Which of the substances Exhibited the greatest molecular forces between water,acetone,methylated spirits and sunflower oil???

Answer 3

To solve this problem, we can use the equations of motion. The key is to analyze the motion of the person in the horizontal and vertical directions separately.

Let's start with the vertical motion. The person jumps from a height h above the lake with an initial vertical velocity of 0 m/s. The only force acting on the person in the vertical direction is gravity (mg), where m is the mass of the person and g is the acceleration due to gravity.

Using the second equation of motion, we have:

h = (1/2)gt^2

Rearranging the equation, we get:

t = sqrt(2h/g)

Next, let's consider the horizontal motion. The person runs horizontally from a distance d behind the edge of the cliff with an acceleration that varies in time according to ax = b1t, where b1 is a constant.

Using the equation of motion in the horizontal direction, we have:

s = ut + (1/2)at^2

Since the person starts from rest (u = 0), the equation simplifies to:

s = (1/2)at^2

Substituting the expression for acceleration ax = b1t, we get:

s = (1/2)b1t^3

Now, we can substitute the expression for t from the vertical motion into the equation for s:

s = (1/2)b1(sqrt(2h/g))^3

Next, we simplify the equation:

s = (1/2)b1(2h/g)sqrt(2h/g)sqrt(2h/g)

s = (b1h/g)sqrt(2h/g)

Therefore, the expression for s in terms of the given quantities d, b1, h, and acceleration due to gravity g is:

s = (b1h/g)sqrt(2h/g)

That is the final answer.