A ball of mass 4g is thrown from a top of a building with a velocity of 20ms^-1 towards another building which is 40m away @ an angle of 30 (above horizontal). @ what distance above or below will the ball strike the second building? (a)4.58m below (b) 3.58m below (c) 3.58m above (d) 4.58m above

Answer 1

Assuming the buildings are the same height,
the horizontal speed is constant at 20cos30° = 17.3 m/s
so, how long will it take to travel the 40m ?
Now recall that the height of the ball (relative to the tops of the buildings) is
h(t) = 20sin30° t - 4.9t^2 = 10t - 4.9t^2

So, using the value of t from part 1, evaluate h(t) to answer part 2.