A particle is projected from point O on the ground with velocity `u = 5(sqrt5) m//s` at angle `alpha = tan^(-1) (0.5)`. It strikes at a point C on a fixed smooth plane AB having inclination of `37^@` with horizontal as shown in figure. If the particle does not rebound, calculate.
(a) coordinates of point C in reference to coordinate system as shown in the figure.
(b) maximum height from the ground to which the particles rises. `(g = 10 m//s^2)`

Answer

A, B, D

Explanation

(a) Let, (x,y) be the coordinates of point C.

`x = OD = OA + AD`
`:. x = 10/3 + y cot 37^@ = (10+4y)/3` ………(i)
As point C lies on the trajectory of a parabola,
we have
`y = x tan alpha - (gx)^2/(2u)^2 (1+tan^2alpha)`.........(ii)
Given that , `tan alpha = 0.5 = 1/2`
Solving Eqs. (i) and (ii), we get x = 5m and y = 1.25m.
Hence, the coordinates of point C are (5m, 1.25m).
(b) Let `v_y` be the vertical component of velocity of
the particle just before collision at C.

Using `v_y = u_y + a_yt, we have `
`v_y = u sin alpha - g (x//ucos alpha) (:' t = x//u cos alpha)`
`= (5sqrt5)/(sqrt5) - (10 xx 5)/(5 (sqrt5) xx 2//(sqrt5)) = 0`
Thus, at C, the particle has only horizontal
component of velocity
`v_x = u cos alpha = 5(sqrt5) xx (2//(sqrt5)) = 10 m//s`
Given, that the particle does not rebound after collision. So, the normal component of velocity
(normal to the plane AB) becomes zero. Now,
the particle slides up the plane due to tangential
component `v_x cos 37^@ = (10)(4/5) = 8m//s`
Let h be the further height raised by the particle.
Then,
`mgh = 1/2 m (8)^2 or h = 3.2 m `
Height of the particle from the ground `= y+h`
`:. H = 1.25 + 3.2 = 4.45m. `