Suppose that 2 j of work is needed to stretch a spring from its natural length of 36 cm to a length of 52 cm. (a) how much work is needed to stretch the spring from 44 cm to 48 cm? (round your answer to two decimal places.) j (b) how far beyond its natural length will a force of 15 n keep the spring stretched? (round your answer one decimal place.) cm

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

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Th answer is 3200 m/s