Question

The surface of a hill is inclined at an angle a to the horizontal. A stone is thrown from the top of the hill at an angle b with the vertical with a velocity v. how far from the top will the stone strike the surface of the hill?

pls answer asap. ​

Answer 1

Solution:-

$Horizontal \: component \: of \: velocity = vcos∅ \\ if \: u \: will \:visualize \: from \: figure \\ u \: can \: understand \: that \\ ∅ = 180 - (90 - \alpha + \beta ) = 90 - ( \alpha - \beta ) \\ so \: horizontal \: component = vcos(90 + \alpha - \beta ) \\ vsin( \alpha - \beta ) \\ this \: will \: be \: the \: velocity \: along \: x \\ similarly \: velocity \: along \: y = vcos( \alpha - \beta ) \\ now \: for \: x \: axis \\ s = vsin ( \alpha - \beta )t - \frac{1}{2}gsin \alpha {t}^{2}......i) \\ for \: y \: axis \\ as \: the \: stone \: will \: again \: fall \: back \: on \: x - axis \\ so \: displacement \: will \: be \: 0 \: along \: y \\ 0 = vcos( \alpha - \beta ) \times t - \frac{1}{2}gcos \alpha {t}^{2} \\ vcos( \alpha - \beta )t = \frac{1}{2} gcos \alpha {t}^{2} \\ t = \frac{2vcos( \alpha - \beta )}{gcos \alpha } \\ now \: putting \: this \: value \: of \: t \: in \: i) \: we \: get \\ s = \frac{2vcos( \alpha - \beta ) \times vsin( \alpha - \beta )}{gcos \alpha } + \frac{1 \times gsin \alpha \times 4 {v}^{2} {cos}^{2}( \alpha - \beta ) }{ {g}^{2} {cos}^{2} \alpha }$

s={2v^2cos(beta-alpha)sinB}/gcos^2a

Note:-here a=∅(taken)

Answer 2

Answer:

Explanation:

grab a look at the attachments