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.

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To find the distance s in terms of the given quantities d, b1, h, and acceleration due to gravity g, we can use the principles of projectile motion.

Let's break down the problem into different segments:

1. Vertical Motion:
- The person jumps off the cliff vertically, experiencing free fall under the influence of gravity.
- The time taken to reach the maximum height can be calculated using the formula: t1 = sqrt((2h) / g).
- The maximum height reached can be calculated using the formula: h_max = (g * t1^2) / 2.

2. Horizontal Motion:
- The person leaves the cliff horizontally, so there is no vertical acceleration.
- The horizontal distance traveled can be calculated using the formula: s = Vx * t2, where Vx is the horizontal component of velocity, and t2 is the total time of flight.
- To find Vx, we need to calculate the acceleration in the horizontal direction. Since ax = b1t, we can integrate this equation to find Vx as a function of time: Vx = (b1 * t^2) / 2.
- The total time of flight can be calculated using the formula: t2 = 2 * t1.

Now, substituting the values we obtained earlier:

t2 = 2 * t1 = 2 * sqrt((2h) / g)
Vx = (b1 * t^2) / 2 = (b1 * (2h / g)) / 2 = (b1h) / g

Finally, substitute Vx and t2 into the equation for s:

s = Vx * t2 = ((b1h) / g) * (2 * sqrt((2h) / g))

Therefore, s = 2 * ((b1h)^2 / g^2) * sqrt((2h) / g)

This gives the value of s in terms of the given quantities d, b1, h, and acceleration due to gravity g.

To find the distance s in terms of the given quantities, we can break down the problem into two parts: the vertical motion and the horizontal motion.

First, let's consider the vertical motion. When the person jumps off the cliff, the only force acting on them is gravity. The vertical motion can be described using the equation:

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

where h is the height of the cliff, g is the acceleration due to gravity, and t is the time taken for the person to reach the surface of the lake.

Next, let's consider the horizontal motion. Since the person leaves the cliff horizontally, the initial horizontal velocity (Vx) is equal to their horizontal speed throughout the jump. The horizontal distance (s) can be calculated using the equation:

s = Vx*t

Now, let's find the horizontal velocity (Vx). We know that the person starts from rest and accelerates with an acceleration that varies with time according to ax = b1t. We can integrate this acceleration function to find the horizontal velocity:

ax = d(Vx/dt)

b1t = d(Vx/dt)

Integrating both sides:

∫(b1t) dt = ∫d(Vx) = d(Vx)

(b1/2)t^2 = dVx

Vx = (b1/2)t^2/d

Now, substitute this value of Vx into the equation for s:

s = Vx*t

s = [(b1/2)t^2/d]*t

s = (b1/2)t^3/d

Finally, substitute the value of t from the equation for vertical motion into the equation for s:

s = (b1/2)*[(2h/g)^(3/2)]/d

Therefore, the distance s in terms of the given quantities is:

s = (b1/2)*[(2h/g)^(3/2)]/d