Let B = 5.05 m at 60.0°. C and A have equal magnitudes. The direction angle of C is larger than that of by 25.0°. Let A · B = 27.6 m and B · C = 33.9 m2. Find the magnitude (in m) and direction (in degrees) of A.

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

Answertechnician (a) is correct;

It is the areas with lines. this may sound a little confusing, but i hope it !

At maximum height, kinetic energy is converted in potential energy  formula: at max height, ke = ½ * 0.27kg * (16m/s) ² = 34.56 j  so the remaining pe = (41.70 – 34.56)= 7.14 j = 0.27 kg * 9.8m/s² * h  = (9.8/7.14 * 0.27)so max height h = 5.08 or 5 m