Question

A particle is projected from point O on the ground with velocity $u=5\sqrt{5}m/s$ at angle $a={\tan }^{-1}\left(0.5\right).$ It strikes at point C on a fixed smooth plane AB having inclination of ${37}^{\circ }$ with horizontal as shown in figure. If the particle does not rebound, calculate coordinates of point C in reference to coordinate system as shown in the figure.

• A. $(12.5m,5m)$
  
• B. $(5m,1.25m)$
  
• C. $(5m,5m)$
  
• D. $(1.25m,1.25m)$
  

Answer 1

The correct option is B $(5m,1.25m)$

If u is the initial velocity, ucosα = horizontal constant velocity (unaffected with vertical acc. g)

usinα = vertical velocity
and t seconds. Let the co-ordinates of point C(a, b)

a = ucosαt ----- (i)

b = usinαt - $\frac{1}{2}$9.81t² ------ (ii)

from equations (i) & (ii) with a substitution for t we get,

b = atanα - $\frac{1}{2}$(9.81)a² / [ucosα]² -------- (iii)

the equation of the line representing the incline which passes through (a, b) is y = xtan37° + c
when y = 0 => x = 3.333, therefore c = -3.333(0.7536) = -2.5118

the equation is y = 0.7536x - 2.5118

therefore b = 0.7536a - 2.5118 --------- (iv)

substituting in equation (iii),

0.7536a - 2.5118 = 0.5a - $\frac{1}{2}$(9.81)a² / [5√5(0.8944)]²

- 2.5118 + 0.2536a = - 4.905a² / 100

251.185 = 4.905a² + 25.36a
a² + 5.17a - 51.21 = 0
on solving the above eqn we get,

solving a ~= 5.02 m, b ~= 1.27 m = (5.02, 1.27) .
Thus the coordinates are (5m,1.25m)