What is the landing location of the projectile and its speed of impact?

A projectile of mass 1 kg is launched from ground level toward the east at 200 m/s, at an angle of $\frac{\pi}{6}$ to the horizontal. If the spinning of the projectile applies a steady northerly Magnus force of 2 newtons to the projectile. ~~~~~~~~~~~ The answer is Landing point: (3534.8,416.5,0) and Speed at impact: 204 m/s, but i don't know how to start the problem.

Feb 15, 2018

$\text{please check math operations.}$

Explanation:

$\text{The projectile will do a three dimensional motion. while}$
$\text{projectile is moving to eastward with horizontal component of }$
$\text{its velocity, the Force of 2N moves it toward north.}$

$\text{The time flight for the projectile is:}$

$t = \frac{2 {v}_{i} \sin \left(\theta\right)}{g}$

$t = \frac{2 \cdot 200 \cdot \sin \left(30\right)}{9.81}$

$t = 20.39 \sec .$

$\text{The horizontal component of initial velocity :}$

${v}_{x} = {v}_{i} \cdot \cos 30 = 200 \cdot \cos 30 = 173.21 \text{ } m {s}^{-} 1$

$\text{x-range :" =v_x*t=173.21*20.39=3531.75 " } m$

$\text{the force with 2N causes a acceleration toward north .}$

$F = m \cdot a$

$2 = 1 \cdot a$

$a = 2 m {s}^{-} 2$

$\text{y_range :} \frac{1}{2} \cdot a \cdot {t}^{2}$

$\text{y-range :} = \frac{1}{2} \cdot 2 \cdot {\left(20.39\right)}^{2}$

$\text{y-range :"=415.75" } m$

$\text{impact velocity:}$

$\text{It falls at a speed of 200 " m s^-1 " in the east direction.}$

${v}_{\text{east}} = 200 m {s}^{-} 1$

v_("north")=a*t=2*20.39=40.78" "ms^-1

$v = \sqrt{{v}_{\text{east")^2+v_("north}}^{2}}$

$v = \sqrt{{200}^{2} + {\left(40.78\right)}^{2}}$

$v = 204.12 \text{ } m {s}^{-} 1$