Correct Answer - (a) 340 m//s
(b) 1870 m.
Here, `t_1 = 4 s and t_2 = 3 s, d_1 = 680 m`
(a) As `d_1 = (vt)/(2), v = (2 d_1)/(t_1) = (2 xx 680)/(4) = 340 m//s`
(b) As `d_2 = (v(t_1 + t_2))/(2) = (340(4 + 3))/(2) = 1190 m`
Distance between two cliffs, `d = d_1 + d_2 = 680 + 1190 = 1870 m`.