# a coin tosses in air with velocity 9.8n m/s ,then what is min height of coin??

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### 2 Answers

Given initial velocity of the coin (u)= 9.8 m/s , The coin will travel in

the air till its velocity becomes 0 assuming no other force except the

gravity is acting on it while it is on the air. Let 'h' be the height of the

coin reached. Its velocity at height 'h' will be=0 i.e the final

velocity(v)=0

Using equation of motion under gravity :

v^2 = u^2 - 2*g*h [ 'g' will be (-) as it is travelling againt the gravity]

=> (0)^2 = (9.8)^2 - 2*9.8*h [ taking value of 'g' = 9.8 m/s^2 ]

=> 2*9.8*h = (9.8)^2

=> h = (9.8*9.8) / (2*9.8)

=> h = 9.8/2 = 4.9

**Hence the height attained by the coin = 4.9 m <-- Answer**

when the coin was tossed I assume that the tosser has pushed mainly at the center of the coin, now really speaking the question is incomplete, because you have not mentioned the radius of the coin and at which part of the coin it was hit, because that could give rotation to the coin too. But as you are saying it moved up with velocity 9.8 m/s (what is 9.8n ?) so I will assume that it rotated very very slowly so that we can ignore the energy lost in the rotation. So by applying energy conservation:

m g h = (1/2) m v^2 + (1/2) I w^2 [m = mass of coin,

g = acceleration due to gravity,

h = height attained by coin

v = initial velocity of coin

I = moment of inertia of coin

which rotates around its diameter

w = rotational speed ~ 0

=> m g h = (1/2) m v^2 [approximately as w~0]

=> h = v^2 / 2 g

= (9.8)^2/(2 9.8)

= 4.9 m

so mean height attained is 4.9 m, approximately.