Start by solving for the height the stone reaches at the top of its trajectory. You can use the kinematic equation Vf^2=Vi^2+2gd where g is the acceleration due to gravity (I used 9.8) and d is the displacement. Knowing that the velocity in the y direction is 0 m/s when the rock is at the max y displacement, you can solve for d. Don't forget that g is negative. This will be the height the rock travels ABOVE the building. It comes out to be something like 1.28 m, so add 30 m and you have part 2.
For part two you should use the kinematic that is d=ViT+0.5GT^2. You'll have to use the quadratic equation to solve for t, but that's OK. Use the answer to part two (add the 30 m) for d and the 5 m/s for Vi. The positive solution is your answer. I got about 2 seconds.
Hope this helps!
John stand at the top of a 30m building. He throws a stone upwards with a speed of 5m/s. Calculate. (1)how long the stone is in the air. (2)the maxi
mum height the stone reaches the ground?
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