A test rocket is launched by accelerating it along a 200.0- incline at 2.38 starting from rest at point A (the figure .) The incline rises at 35.0 above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored).



The greatest horizontal range of the rocket beyond point A.
(m)

200 degrees incline? That's strange... I'll take it as 20 degrees from the horizontal.

Make a sketch first.

The speed along the slope is given by:

v^2 =u^2 + 2as

where v is the final speed,
u the initial speed
a the acceleration
s the displacement.

From your sketch, we get the displacement to be 35/sin(20) m = 102.3 m

Hence, the speed of the rocket at the end of the slope is:

v^2 = 0 + 2(2.38)(102.3)

v = 15.6 m/s

Now, at the top of the slope, there is another equation to use after breaking the velocity of the rocket into two components.

Vertical component is given by:

v_v = 15.6sin20 = 5.34 m/s

v_h = 15.6 cos20 = 14.7 m/s

The vertical velocity of the rocket is given by:

s_v= u_vt + \frac12 at^2

The horizontal velocity given by:

s_h = u_ht

Set sv to -35 to get the time of flight.

Use this time to get the horizontal displacement from the launching point.

Use trigonometry to get the horizontal displacement while the rocket travels along the incline.

Add both horizontal displacements to get the greatest range.

Post what you get!