assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves the uzzel of t

Question

assume that the initial speed is 25 m/s and the angle of projection is 53 degree above the hroizontal. the cannon ball leaves the uzzel of the cannon at a highet of 200 m.( the cannon is at the edge of the cliff) A: find the horizontal distance the cannon travles. B: when does the cannon ball reach the ground? C: find the maximum highet the cannon ball reaches.

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Sigridomena 4 years 2021-08-27T19:07:23+00:00 1 Answers 16 views 0

Answers ( )

  1. Farah
    0
    2021-08-27T19:08:53+00:00
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    Answer:

    A.  xmax = 131.49 m

    B.  t = 8.74 s

    C.  ymax = 220.33 m

    Explanation:

    A. In order to find the horizontal distance which cannon travels you first calculate the flight time. The flight time can be calculated by using the following formula:

    y=y_o+v_osin\theta-\frac{1}{2}gt^2      (1)

    yo: height from the projectile is fired = 200m

    vo: initial velocity of the projectile = 25m/s

    g: gravitational acceleration = 9.8 m/s^2

    θ: angle between the direction of the initial motion of the ball and the horizontal = 53°

    t: time

    You need the value of t when the projectile hits the ground. Then, in th equation (1) you make y = 0m.

    When you replace the values of all parameters in the equation (1), you obtain the following quadratic formula:

    0=200+(25)sin53\°t-\frac{1}{2}(9.8)t^2\\\\0=200+19.96t-4.9t^2 (2)

    You use the quadratic formula to obtain the value of t:

    t_{1,2}=\frac{-19.96\pm\sqrt{(19.96)^2-4(-4.9)(200)}}{2(-4.9)}\\\\t_{1,2}=\frac{-19.96\pm65.71}{-9.8}\\\\t_1=8.74s\\\\t_2=-4.66s

    You use the positive value because it has physical meaning.

    Now, you can calculate the horizontal range of the projectile by using the following formula:

    x_{max}=v_ocos\theta t      

    x_{max}=(25m/s)(cos53\°)(8.74s)=131.49m

    The cannon ball travels a horizontal distance of 131.49 m

    B. The cannon ball reaches the canon for t = 8.74s

    C. The maximum height is obtained by using the following formula:

    y_{max}=y_o+\frac{v_o^2sin^2\theta}{2g}     (3)

    By replacing in the equation (3) the values of all parameters you obtain:

    y_{max}=200m+\frac{(25m/s)^2(sin53\°)^2}{2(9.8m/s^2)}\\\\y_{mac}=200m+20.33m=220.33m

    The maximum height reached by the cannon ball is 220.33m

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