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 - 129.81t² ------ (ii)
from equations (i) & (ii) with a substitution for t we get,
b = atanα - 12(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 - 12(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)