49. Using Eq. 4-26, the launch speed of the projectile is gR (9.8 m/s 2 )(69 m) v0 = = = 26.52 m/s . sin 2θ sin 2(53°)
The horizontal and vertical components of the speed are vx = v0 cos θ = (26.52 m/s) cos 53° = 15.96 m/s v y = v0 sin θ = (26.52 m/s) sin 53° = 21.18 m/s.
Since the acceleration is constant, we can use Eq. 2-16 to analyze the motion. The component of the acceleration in the horizontal direction is ax =
vx2 (15.96 m/s) 2 = = 40.7 m/s 2 , 2 x 2(5.2 m) cos 53°
and the force component is Fx = max = (85 kg)(40.7 m/s 2 ) = 3460 N. Similarly, in the vertical direction, we have v y2 (21.18 m/s) 2 ay = = = 54.0 m/s 2 . 2 y 2(5.2 m) sin 53° and the force component is Fy = ma y + mg = (85 kg)(54.0 m/s 2 + 9.80 m/s 2 ) = 5424 N.
Thus, the magnitude of the force is F = Fx2 + Fy2 = (3460 N) 2 + (5424 N) 2 = 6434 N ≈ 6.4 × 103 N, to two significant figures.
