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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:
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*tv-u = a*t
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 = 0.0204 seconds
Hence the total time taken for coin to reach back on the ground is 0.204 + 0.204 = 0.408 secondsPlease let me know if you need any clarification. I'm always happy to answer your questions. Hope you find my answer satisfactory. Rate me 5. :)
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