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

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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 $\frac{π}{6}$ $pi/6$ to the horizontal. If the spinning of the projectile applies a steady northerly Magnus force of 2 newtons to the projectile.

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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.

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Answer 1 · 2018-02-15T18:50:03

$please check math operations.$ $"please check math operations."$

Explanation:

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

$The time flight for the projectile is:$ $"The time flight for the projectile is:"$

$t = \frac{2 v_{i} sin (θ)}{g}$ $t=(2 v_i sin(theta))/g$

$t = \frac{2 \cdot 200 \cdot sin (30)}{9.81}$ $t=(2*200*sin (30))/(9.81)$

$t = 20.39 sec .$ $t=20.39 sec.$

$The horizontal component of initial velocity :$ $"The horizontal component of initial velocity :"$

$v_{x} = v_{i} \cdot cos 30 = 200 \cdot cos 30 = 173.21 m s^{- 1}$ $v_x=v_i*cos 30=200*cos 30=173.21 " "ms^-1$

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

$the force with 2N causes a acceleration toward north .$ $"the force with 2N causes a acceleration toward north ."$

$F = m \cdot a$ $F=m*a$

$2 = 1 \cdot a$ $2=1*a$

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

$y_range : \frac{1}{2} \cdot a \cdot t^{2}$ $"y_range :" 1/2*a*t^2$

$y-range : = \frac{1}{2} \cdot 2 \cdot {(20.39)}^{2}$ $"y-range :"=1/2*2*(20.39)^2$

$y-range : = 415.75 m$ $"y-range :"=415.75" "m$

$impact velocity:$ $"impact velocity:"$

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

$v_{east} = 200 m s^{- 1}$ $v_("east")=200 ms^-1$

$v_{north} = a \cdot t = 2 \cdot 20.39 = 40.78 m s^{- 1}$ $v_("north")=a*t=2*20.39=40.78" "ms^-1$

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

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

$v = 204.12 m s^{- 1}$ $v=204.12 " "ms^-1$

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

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