Question
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
Answers
oobleck
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.
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.