A body falling freely from a given height H hits on inclined plane in

1.	A body falling freely from a given height H hits on inclined plane in its path at a height h. As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of h/H the body will take maximum time to reach the ground?
Answer» HEIGHT from which the ball is released from rest is H height upto which it MOVES vertically downward is (H-h) LET the time upto which,the ball fall vertically diwnward be t1 and time in which it covers parbolic path be t2 using first equation of motion t1={2(H-h)/G}^1/2 similarlly t2=(2h/g)^1/2 T={2(H-h)/g}^1/2+{2h/g}^1/2. (adding both,to get the total time to REACH the ground) now,for maximum time,to reach the ground dT/dH=0 On differenting it we get h/H=1/2

A body falling freely from a given height H hits on inclined plane in its path at a height h. As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of h/H the body will take maximum time to reach the ground?

Answer»

HEIGHT from which the ball is released from rest is H

height upto which it MOVES vertically downward is (H-h)

LET the time upto which,the ball fall vertically diwnward be t1 and time in which it covers parbolic path be t2

using first equation of motion

t1={2(H-h)/G}^1/2

similarlly

t2=(2h/g)^1/2

T={2(H-h)/g}^1/2+{2h/g}^1/2. (adding both,to get the total time to REACH the ground)

now,for maximum time,to reach the ground

dT/dH=0

On differenting it we get

h/H=1/2

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A body falling freely from a given height H hits on inclined plane in its path at a height h. As a result of this impact the direction of the velocity of the body becomes horizontal. For what value of h/H the body will take maximum time to reach the ground?

LET the time upto which,the ball fall vertically diwnward be t1 and time in which it covers parbolic path be t2

using first equation of motion

t1={2(H-h)/G}^1/2

similarlly

t2=(2h/g)^1/2

T={2(H-h)/g}^1/2+{2h/g}^1/2. (adding both,to get the total time to REACH the ground)

now,for maximum time,to reach the ground

dT/dH=0

On differenting it we get

h/H=1/2

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