Physics, 25.11.2020 14:00 SavannahP04

An elevator, in which a man is standing moving upward with a constant speed of 10m/sec2. If a man drops a coin from a height of 2.5m. Find the time
taken by it to reach the oor of the elevation? g=9.8m/sec2

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Answer from: davestrider404

We are Given:

velocity of the lift (u₁) = 10 m/s

initial velocity of the coin (u₂) = 10 m/s

(the coin is also moving with the Elevator)

acceleration of the coin (a₂) = -9.8 m/s²

acceleration of the Elevator (a₁) = 0 m/s

Distance covered by the coin (s) = -2.5 m

Relative Velocity and Acceleration:

Relative velocity:

We will find the velocity and acceleration of the coin with respect to the lift since we are monitoring the motion of the coin

velocity of the coin with respect to the elevator (u₂₁) = u₂ - u₁

u₂₁ = 10 - 10

u₂₁ = 0m/s

Relative Acceleration:

acceleration of the coin with respect to the elevator (a₂₁) = a₂ - a₁

a₂₁ = -9.8 - 0

a₂₁ = -9.8

Solving for the Time taken:

From the second equation of motion, we know that:

s= ut + 1/2at₂

we can rewrite this equation in terms of relative motion:

s = u₂₁(t) + 1/2(a₁₂)t²

Notice that the time and the displacement are not relative, that's because displacement and time will remain the same no matter the frame of reference

replacing the known values in the equation:

-2.5 = (0)(t) + 1/2 (-9.8)(t²)

-2.5 = -4.9(t²)

dividing both sides by -4.9

t² = -2.5 / -4.9

t² = 25/49

t² = (5)² / (7)²

taking the square root of both the sides

t = 5/7 OR 0.71 seconds (approx)

Therefore, the coin will reach the floor of the Elevator in 0.71 seconds

Answer from: Quest

If football was tossed in air without gravity or friction the ball would go into space

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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