1.

A particle is projected from a point on the level ground and its height is `h` when at horizontal distances `a and 2 a` from its point of projection. Find the velocity of projection.

Answer» Correct Answer - `(1)/(2) sqrt(((4 a^2)/(h) + 9 h)) g`.
If `v_0` is the velocity of projection and `prop` the angle of projection, the equation of trajectory is
`y = x tan prop - (1)/(2) (gx^2)/(v_0^2 cos^2 prop)`…(i)
With origin at the point of projection,
`gx^2 -m 2 v_0^2 sin prop cos prop. x + 2 v_0^2 cos^2 prop.y = 0` ...(ii)
Since the projectile passes through two points `(a,h) and (2a, h)`, then `a and 2a` must be roots of Eq. (ii),
`a + 2 a = (2 v_0^2 sin prop cos prop)/(g)`...(iii)
and `a xx 2a = (2 v_0^2 cos^2 prop h)/(g)`...(iv)
Dividing Eqs (iii) by (iv), we get
`(3a)/(2a^2) = (tan prop)/(h) or tan prop = (3 h)/(2a)`
From Eq. (iv),
`v_0^2 = (ga^2)/(h) sec^2 prop = (ga^2)/(h) (1 + tan^2 prop) = (ga^2)/(h) (1 + (9 h^2)/(4 a^2))`
=`(g)/(4) ((4a^2)/(h) + 9h) or v_0 = (1)/(2)sqrt((4a^2)/(h) + 9 h)) g`.


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