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If you throw a coin in the air, it will trace out a parabola with acceleration -9.8 m/s^2 (gravity on earth).  In other words, its height is quadratic as a function of time,

h(t)= at^2+bt+c

Suppose that you throw a coin directly upward from ground level at 2 m/s.  How long will it take to return to the ground?

Oct 14th, 2015

The following equation of motion can be used to solve any problem where the acceleration is constant and the direction of motion is also constant.

v = u + a*t

Here, 'v' is the initial velocity, 'u' is the final velocity, a is the acceleration and t is the time taken.

Here it is given that the acceleration is 9.8m/s, but the obviously, the direction changes once, i.e. at the top, when it stops and starts falling back.

So let us divide the problem into two parts:

a)From ground level to maximum height:

u=2 m/s

v=0 m/s ( Speed is momentarily zero at maximum height, as the direction of velocity reverses)

a= -9.8 m/s^2 (negative acceleration as the slows down)

t=?  (to be calculated)

using the equation

v = u + a*t

v-u = a*t

t= (v-u)/a

t=(2-0)/9.8 = 0.204 seconds

b) From maximum height to reaching back on ground:

By principle of conservation of energy, an object thrown up vertically will have the same when it reaches back at the same height, as it has the same Kinetic Energy.

v= 2 m/s

u = 0 m/s ( Speed is momentarily zero at maximum height, as the direction of velocity reverses)

a = 9.8 m/s ( Positive acceleration as the speed increases when it falls down)

Applying the same equation, v=u+a*t

t=(v-u)/a

t = 0.0204 seconds

Hence the total time taken for coin to reach back on the ground is 0.204 + 0.204 = 0.408 seconds

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Oct 14th, 2015

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Oct 14th, 2015
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Oct 14th, 2015
Sep 23rd, 2017
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