If the acceleration of the projective is: a = c s m/s 2 Where c is a constant that depends on the initial gas pressure behind the projectile

Question

If the acceleration of the projective is: a = c s m/s 2 Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s=1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with a velocity of 200 m/s?

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Neala 4 years 2021-09-05T12:17:35+00:00 1 Answers 9 views 0

Answers ( )

  1. thienhuong
    0
    2021-09-05T12:18:58+00:00
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    Answer:

    c = 4,444.44

    Explanation:

    You have the following expression for the acceleration of the projectile:

    a=cs   (1)

    s: distance to the ground of the projectile

    To find the value of the constant c you use the following formula:

    v^2=v_o^2+2a \Delta s   (2)

    vo: initial  velocity = 0 m/s

    v: final speed = 200 m/s

    Δs: distance traveled by the projectile = 3m – 1.5m = 1.5m

    You replace the expression (1) into the expression (2):

    v^2=2(cs)\Delta s

    You do the constant c in the last equation, then you replace the values of v, s and Δs:

    c=\frac{v^2}{2s\Delta s}=\frac{(200m/s)^2}{2(3m/s^2)(1.5m)}=4444.44

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