A ball which weigh's about 25 kg is subjected to an average horizontal force of 300 N. The ball is stationed on a platform 10 km from the ground, determine the range of the projectile if the impact time of the subjected force is 2 s. The range of the projectile is measured from the end of the platform A to where the projectile strikes the ground at B.

1) 1.08 km
2) 1.54 km
3) 2.55 km
4) 2.49 km

Applying the equation of motion to the ball we have
that
m(v_x)₁ + S F_x dt = m(v_x)²
25(0) + 300 (2) = 25v
v = 24 m/s

Considering the vertical motion, the vertical component of initial veloctiy is (v_o)_y = 0. The initial and final position is (s_o)_y = 0 and s_y = -10km, respectively.
s_y = (s_o)y + (v_o)_y t + 1/2 (a_c)_y t²
-10 (10³) = 0 + 0 + 1/2 (-9.81) t²
t = 45.15 s

By considering the horizontal motion, the horizontal component of velocity is (v_o)_x = 24 m/s. The initial and final position are (s_o)_x = 0 and s_x
s_x = (s_o)_x + (v_o)_x t
R = 0 + 24(45.15) = 1083.65 m
R = 1.083 km

So the range of the projectile is R = 1.08 km