Initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, hβ = 245 m final height, hβ = 0a = -9.8 m/sΒ²time taken = t secondss = ut + 0.5atΒ²β (hβ-hβ) = ut + 0.5atΒ²β 0-245 = 4.9t + 0.5Γ(-9.8)ΓtΒ²β -245 = 4.9t - 4.9tΒ²β 4.9tΒ² -4.9t -245 =0solving it, we get t = 7.59sv = u + at = 4.9 -9.8Γ7.59 = 4.9 - 74.38 = -69.48 m/svelocity is 69.48 m/s downward