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Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward

The boat's velocity relative to the riverbank can be resolved into two components. the component going north is 2.7 m/s and the component going west is 1.2 m/s.we add these two vectors and then find the magnitude of the resultant to get the speed.the magnitude of the resultant vector from adding can be obtained by the pythagorean theorem.    to two significant figures, the boat's speed relative to the riverbank is 3.0 m/s.