Question

A body is projected horizontally from certain height.After sqrt(3) seconds its direction of motio makes 60^(@) with the horizontal.Then its initial velocity of projection is (in ms^(-1) )​

Answer 1

Answer:

After ./3 s the vertical velocity will be got by using v = u + g t

Here u = 0, t = ./3

So v = ./3 g m/s ----(1)

Let V be the velocity after ./3 sec

Then as it inclines at 60 deg with horizontal its horizontal component = V cos 60 = 1/2 * V

Vertical component = V sin 60 ---(2)

(1) and (2) represent the same

So equating V sin 60 = ./3 g

Or V = 2 g ( since sin 60 = ./3 /2)

You are sure that the horizontal component always remain the same.  

Hence the horizontal velocity of projection = 1/2 * 2g = g m/s

Step-by-step explanation:

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Answer 2

Answer:

hope it will help you mate