projectile....

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a particle is projected with a velocity 2âˆšgh so tht it just clears two walls of equal height h which r at a distance 2h frm each other. show that the time of passing b/w the walls is 2âˆšh/g.....

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Lokesh Verma ·2009-11-10 00:00:38

Hint:

Find the velocity of the projectile when it just reaches the top of one of the walls
After which it becomes a projectile with range of 2h

Now can you solve it?

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fibonacci ·2009-11-10 01:06:23

assume initial projection angle=Î¸ , g=10m/s²
so at first wall you have
v_x=√40hcosÎ¸
v_y=√40h²sin²Î¸-20h
so velocity at 1st wall=√40h²-20h
let angle be Ï†
we have \frac{(40h^2-20h)sin2\phi }{g}=2h
put sin2\phi=2cos^2\phi -1 and get cos\phi =\sqrt{\frac{h}{2h-1}}
so horizontal velocity at 1st wall=√20h=√2gh
distance to travel = 2h
divide distance by speed to get the result

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projectile....

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