1. The problem statement, all variables and given/known data A brick is dropped from a hot air ballon which is ascending at 5ms^-1. The height of the balloon above the ground at the time of release of the brick was 30m. a. Determine the time that it took the brick to hit the ground. b. At what height was the balloon above the ground at the instant the brick hit the ground? c. What was the velocity of the brick when it hit the ground? d. Sketch a graph of the velocity against time for the brick over this journey. 2. Relevant equations 1. v=u+at 2. v^2=u^2+2as 3. s=ut+1/2at^2 4. v=s/t 3. The attempt at a solution For this question I specified up direction as negative and down direction as positive. For (a) I decided to use the formula, s=ut+1/2at^2, where I substituted s for 30, u for -5 and a for 9.8. When rearranged this gave me 4.9t^2-5t-30=0. Using the quadratic formula I then got the answer t=-2.53 and 2.53. I would have thought that these values would have been different however as after being released from the balloon, the brick would surely have risen slightly before gravity accelerates it towards ground meaning the positive value should be greater than the negative value. For (b) I used the s=ut+1/2at^2 formula with s=30, u=-5 and a=0 values of the ballon to prove that after six seconds the balloon was 30m above the ground. I got a t=-6 answer for this though and thus maybe I did something wrong, it should be positive right? The I used the s=ut+1/2at^2 formula using -8.53 (-6-2.53)*-5 to give me the displacement 42.65m. For (c) I used v=u+at with the values, u=-5 a=9.8 and t=-2.53 to give me the velocity 19.794 ms^-1 down. I will attempt (d) after finishing a,b and c.