Answer
(a) $v_x = 13 ~m/s$
$v_y = 12 ~m/s$
(b) The ball will strike the building 36 meters below the top.
Work Step by Step
(a) $v_x = v ~cos(\theta) = (18 ~m/s) ~cos(42^{\circ}) = 13 ~m/s$
$v_y = v ~sin(\theta) = (18 ~m/s) ~sin(42^{\circ}) = 12 ~m/s$
(b) $t = \frac{x}{v_x} = \frac{55 ~m}{13 ~m/s} = 4.2 ~s$
We can use the time t to find the vertical displacement:
$y = v_{y0}t + \frac{1}{2}at^2$
$y = (12 ~m/s)(4.2 ~s) + \frac{1}{2}(-9.80 ~m/s^2)(4.2 ~s)^2$
$y = -36 ~m$
The ball will strike the building 36 meters below the top.
