A stone is thrown vertically upward. On its way up it point A with speed v,and point B , 3m higher than A, with speed 1/2 v. Calculate (a) the speed v (b) the maximum height reached by the stone above point B.
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As firstly we are writing equation when stone goes from A TO B
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 = 1 m
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 = 1 m
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