A-particle-is-projected-horizontally-with-speed-u-from-point-A-which-is-10-m-above-the-ground-If-the-particle-hits-the-inclined-plane-perpendicularly-at-point-B-g-10-m-s-2-Find-horizontal-spee

Question Number 16044 by Tinkutara last updated on 17/Jun/17

A particle is projected horizontally

with speed u from point A, which is 10

m above the ground . If the particle hits

the inclined plane perpendicularly at

point B . [g = 10 m / s^{2}]

Find horizontal speed with which the

particle was projected .

Commented by Tinkutara last updated on 17/Jun/17

Answered by ajfour last updated on 17/Jun/17

A s t h e p r o j e c t i l e h i t s a 45 ° p l a n e

n o r m a l l y, i t s v e l o c i t y t h e r e (a t B)

i s d i r e c t e d a t a n a n g l e o f 45 °

b e l o w t h e h o r i z o n t a l .

{(v_{y})}_{a t B} = - {(v_{x})}_{a t B}

- g t = - u \Rightarrow t = \frac{u}{g}

a l s o i f o r i g i n i s t a k e n a t t h e

t h e s t a r t o f i n c l i n e,

a t p o i n t B, y = x

h - \frac{1}{2} {g t}^{2} = u t, w i t h t = u / g

w e h a v e

h - \frac{g}{2} (\frac{u^{2}}{g^{2}}) = \frac{u^{2}}{g}

h = \frac{u^{2}}{2 g} + \frac{u^{2}}{g}

\Rightarrow \frac{3 u^{2}}{2 g} = h

o r u = \sqrt{\frac{2 g h}{3}} = \sqrt{\frac{2 \times 10 \times 10}{3}} m / s

= 10 \sqrt{\frac{2}{3}} m / s .

Commented by Tinkutara last updated on 17/Jun/17

Thanks Sir!