Recollect the past 6!
A Shot is fired with a velocity u at a vertical wall whose distance from the point of projection is x . the greatest height above the level of the point of projection at which the bullet can hit the wall is
Details And Assumptions
g= 10 SI unit
x = 4 m
u = 10 m/sec
For the problem writing party
The answer is 4.2.
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1 solution
We know, H = x t a n A − g ( x 2 s e c 2 A / u 2 ) where A is the angle of projection.
d H / d A = x s e c 2 A ( 1 − x g t a n A / u 2 )
For Maximum, d H / d A = 0 Therefore we obtain, 1 − g x t a n A / u 2 = 0 or t a n A = u 2 / g x Substituting this value of t a n A in our trajectory equation gives, H = ( u 4 − g 2 x 2 ) / 2 u 2 g
This is the maximum height attained by the bullet.
Putting reqiured values gives H = 4 . 2 m .