A person, standing on a vertical cliff a height h above a 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 bottom of the cliff. 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 rest, runs at a constant acceleration a and then leaves the cliff horizontally. The person just clears the rock. Find s in terms of the given quantities d , a , h , and the gravitational acceleration g

Question

A person, standing on a vertical cliff a height h above a 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 bottom of the cliff. 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 rest, runs at a constant acceleration a and then leaves the cliff horizontally. The person just clears the rock. Find s in terms of the given quantities d , a , h , and the gravitational acceleration g

Answer 1

0.4.9*t^2 = h

t^2 = h/4.9
t = sqrt (h)/2.21 = Fall time.

S = Vx * t

Answer 2

To find the distance s in terms of the given quantities d, a, h, and the gravitational acceleration g, we can use the equations of motion.

Let's break down the problem and find the time it takes for the person to reach the edge of the cliff. The person starts from rest and accelerates at a constant acceleration a, so the equation for the distance traveled in terms of time is:

s = (1/2) * a * t^2 ---(1)

where t is the time taken to reach the edge of the cliff.

Now, let's find the time it takes for the person to reach the edge of the cliff.

The person starts from rest, so the initial velocity u is 0. The final velocity v can be calculated using the equation:

v = u + a * t

Since u is 0, the equation simplifies to:

v = a * t ---(2)

Next, let's consider the vertical motion of the person when they jump off the cliff.

The person falls downward, so the equation for the vertical distance fallen in terms of time is:

h = (1/2) * g * t^2 ---(3)

where g is the gravitational acceleration.

Since the person just clears the rock, the vertical distance fallen is h - s.

Now, we have two equations: equation (1) for horizontal motion and equation (3) for vertical motion. By solving these two equations simultaneously, we can find the value of time t.

From equation (1), we can rearrange it to express t in terms of s:

t = sqrt((2s)/a) ---(4)

Substituting this value of t into equation (3), we have:

h = (1/2) * g * ((2s)/a) ---(5)

Now, we can solve equation (5) for s:

Multiply both sides of the equation by a:

2ha = g * 2s

Divide both sides of the equation by g:

s = (ha) / g

Therefore, the distance s in terms of the given quantities d, a, h, and g is:

s = (ha) / g

This equation gives us the value of s in terms of d, a, h, and g, and can be used to determine the distance s needed to just clear the rock when jumping off the cliff.