Question

A particle is projected from a horizontal plane to pass over two objects at heights h and k and a slant distance d apart. The least possible speed of projection is :-

(a) g ( h + k + d )

(b) √g ( h + k + d )

(c) h ( g + k + d )

(d) √ h ( g + h + d )

Answer 1

the range of a projectile is 2*u^2*sin(Theta)*cos(Theta)/g. This is twice the maximum height.

In both the above relations theta is unknown, we need to find that out.

2*Max height = Range

hence:

u^2*sin(Theta)/g = 2*u^2*sin(Theta)*cos(Theta)/g

Solving this out we get the angle of projection, which is given as follows:

tan(Theta) = 2, which can be further solved to get sin(Theta) = 2/Sqrt(5) and cos(Theta) = 1/Sqrt(5)

Put this in the formula of range and we get range = 4u^2/(5g)

Answer 2

Answer:

Explanation:

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