A missile is moving 1350 m/s at a 25.0 degree angle. It needs to hit a target 23500 m away in a 55.0 degree direction in 10.20s. What is t

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A missile is moving 1350 m/s at a 25.0 degree angle. It needs to hit a target 23500 m away in a 55.0 degree direction in 10.20s. What is the direction of its final velocity?

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Mộc Miên 5 years 2021-09-03T10:30:29+00:00 1 Answers 67 views 0

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  1. minhkhang
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    2021-09-03T10:31:45+00:00
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    Answer:

    V=3204m/s

    Direction is normal to the incline

    Explanation:

    This question tests on projectile motion.

    First, calculate the acceleration of the missile.

    Incline=55º,Distance=23500m, t=10seconds.

    Horizontal motion of projectile

    x=V_xt+0.5a_xt\\23000Cos55\textdegree=1350Cos25\textdegree\times10.20+0.5a_x(10.20)^2\\a_x=19.2m/s^2

    Final velocity

    (V_x)\prime=V_x+a_xt=1350Cos25\textdegree+19.2\times10.20=1419m/s

    The vertical motion of the missile can be calculated as:

    23500 Sin55.0\textdegree =1350Sin25\textdegree\times10.2+0.5a_y\times10.2^2\\a_y=258.2m/s^2

    Final Velocity is:

    (V_y)\prime=V_y+a_yt=1350Sin25\textdegree+258.2\times10.20=3204m/s

    Combining both we get

    V=\sqrt(V\prime x^2+V\primey^2)=3504.2m/s

    *Misile’s motion is normal to the 55º incline.

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