Question

A stone is thrown vertically upward. On its way up it passes
point A with speed of v, and point B, 3 m higher than A, with
speed v/2. The maximum height reached by stone above point
Bis​

Answer 1

Answer:

down

Explanation:

At A its velocity is v, and at B its velocity is v/2 and their separation is 3 meter 

using formula of   v2 - u2 =2as

{v2}2 − v2   =2(−g) × 3v24 −v2 =−6g−3v24 =−6gv2 = 8g 

Again writing same equation from B to C 

02 - (v/2)2    = -2gh                  (let h is the height from B to highest point)

-v2/4  = -2gh

​v2  = 8gh 

8g = 8gh                                  (putting the value of  v2)

​h = 8g/8g  = 1m

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