4. Tim is walking down the aisle of a speeding bus and tosses a coin in the air. Matthew is standing on the sidewalk outside.

a. Draw the trajectory of the coin from Tim's frame of reference.

b. Draw the trajectory of the coin from Matthew's frame of reference.

What are you trying to ask?

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

Make them both to where they have the denominator of 18that makes 3/2 equal 27/18 and 1/9 equal 2/18so now you have  27/18x+5=2/18subtract 5 from the other sideso now you have 27/18x+2/18-5then subtract 2/18 from 27/18 to get 25/18x=-5then divide 25/18 by -5 and your answer will be  -5/18