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A stone is thrown at a speed of 20 m//s ...

A stone is thrown at a speed of `20 m//s` from top of a building of height 25 m. Find out how far from the foot of building can the stone be thrown.

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Verified by Experts

This can be maximizing x varying `theta` on ground.

By `y=x tan theta -(gx^(2)(1+tan^(2) theta))/(2u^(2))`
`-25=x tan theta-(10x^(2)(1+tan^(2) theta))/(2xx400)`
`tan^(2)theta(x^(2))-tan(80x)+(x^(2)-2000)=0`
It is quadratic in `tan theta` & for `tan theta` to be real
`6400x^(2)-4x^(2)(x^(2)-2000) gt 0`
`x_("max")=60 m`
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