2. A body projected vertically up crosses points A and B separated by 28 m with velocities one-third and one-fourth of the initial velocity, respectively. What is themaximum height reached by it above the ground?

2. A body projected vertically up crosses points A and B separated b

1.	2. A body projected vertically up crosses points A and B separated by 28 m with velocities one-third and one-fourth of the initial velocity, respectively. What is themaximum height reached by it above the ground?
Answer» ANSWERLET initial velocity of the body = uVelocity of the body at POINT A = 3u Velocity of the body at point B = 4u Apply v 2 −u 2 = 2aS for points A and B:( 4u ) 2 −( 3u ) 2 =2(−g)×28On SOLVING we GET: u 2 = 1152gNow apply direct formula for maximum HEIGHTH max = 2gu 2 H = 2g1152g = 576 m

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2. A body projected vertically up crosses points A and B separated by 28 m with velocities one-third and one-fourth of the initial velocity, respectively. What is themaximum height reached by it above the ground?​

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2. A body projected vertically up crosses points A and B separated by 28 m with velocities one-third and one-fourth of the initial velocity, respectively. What is themaximum height reached by it above the ground?