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chaylaceyx3

  • one year ago

A projectile is fired from the ground at an initial angle of 36.4 degrees above the horizontal and an initial muzzle speed of 23.5 m/s. There is also a wall a distance of 41.5 m away from the launcher. A) at what time will the projectile hit the wall? B) how high on the wall will the projectile hit? C) is the projectile traveling upwards or downwards when it hits the wall? How do you know?

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  1. anonymous
    • one year ago
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    |dw:1444327220544:dw|

  2. anonymous
    • one year ago
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    vcos@ = 23.5 * cos(36.4)' = 23.5 * 0.8049 = 18.9 m/s 41.5/ 18.9 = 2.19 s time taken = 2.19 s\[s=ut +(1/2) at ^{2} = 13.94 * 2.19 + 1/2 * 9.8 * 2.19*2.19\] s= 7 m \[H _{MAX}=\frac{v ^{2}\sin ^{2}@ }{ 2g } \] sin 36.5' = .5934 v sin @ = 23.5 * 0.5934 =13.94 m/s*******************************\[13.94^{2}= 194.32\] 194.32/ 19.6 = 9.9 m = height maximum

  3. anonymous
    • one year ago
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    @chaylaceyx3

  4. anonymous
    • one year ago
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    as ht attained before hitting the wall is less than max ht. thus it hits the wall traveling upward.

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