If you exert 250j of work onto a lever that is 5m long, what amount of force are you applying?

From the formula of W = F·d , becuase we have the values for W and d we can find F

W = F·d 
F= W/d
  = 250/5
  = 50 N

40 N of force was applied

We can't tell from the information you gave. Work = (force) times (distance), but that's the distance the force moves. It makes no difference how long the lever is, only how far the force moves.

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