On level ground a shell is fired with an initial velocity of 50.0 m/s at 60.0 degrees above the horizontal and feels no appreciable air resistance. a.) find the horizontal and vertical component's of the shells initial velocity. b.)how long it take the shell to reach its highest point? . c.) find its maximum height above the ground. d.) how far from its firing point does the shell land? . e.) at its highest point, find the horizontal and vertical components of its acceleration and velocity.

A) The two components can be found using trigonometry:
v_x = v · cosα = 50 · cos(60) = 25 m/s
v_y = v · sinα = 50 · sin(60) = 43.3 m/s

b) We now consider only the vertical component of the motion, which is a decelerated motion; at its highest point the final velocity v_f is zero, therefore the time can be found with the inverse formula of the definition of acceleration, a = (v_f - v_i)/t
which in our case is: 
t = (v_f - v_y) / a
where a = -g = -9.8m/s² (g is directed towards the ground, since the shell is going up we have to put the sign minus)
Plugging in numbers:
t = -43.3 / (-9.8) = 4.42 s

c) The maximum height is the displacement after the time needed to reach the highest point, that is:
h = v_y · t + 1/2· a · t² = 43.3 × 4.42 + 1/2 × (-9.8) × (4.42)² = 95.7 m

d) Now we are considering the horizontal component of the motion; the time the shell has to move horizontally is equal to the time taken to reach the maximum height and fall back. The time to free fall from the highest point to the ground is equal to the time needed to reach the highest point from the ground. Therefore, the time allocated for the motion is t = 8.84 s
The horizontal motion is a simple constant motion, therefore:
x = v_x · t = 25 × 8.84 = 221m

e) Vertical components: at the highest point, as we said before, the shell is stationary and ready to fall down. Therefore v_y = 0m/s, while a_y = +9.8m/s².
Horizontal components: at the highest point, again as we said before, the shell is in a constant motion, therefore: v_x = 25m/s and a_x = 0m/s².

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