(a) The vertical motion is accelerated by gravity. The horizontal component is constant (neglecting air resistance, if this is not a projectile motion, the horizontal component would also be accelerated)
(b) Vertical:
[tex]v_y = 30\sin 40^\circ = 19.3\frac{m}{s}[/tex]
Horizontal:
[tex]v_x = 30\frac{m}{s}\cos 40^\circ = 23.0 \frac{m}{s}[/tex]
(c) Use the kinematic equation for distance. Calculate only the vertical component (horizontal is irrelevant):
[tex]s_y = -\frac{1}{2}gt^2 +vt\,,\,\,\,s_y=0\\0 = -\frac{1}{2}gt^2 +vt\\\frac{1}{2}gt =v\\t = \frac{2v}{g}= \frac{2\cdot30\sin 40^\circ\frac{m}{s}}{9.8\frac{m}{s^2}}=3.9s[/tex]
The ball will be in the air for about 3.9 s.
(d) The range is the horizontal distance traveled. We know the ball is in the air for 3.9s and it moves with a horizontal velocity of 23 m/s. So:
[tex]s_x = 23\frac{m}{s}\cdot 3.9 s = 89.7m[/tex]
The range is 89.7 meters.
