Projectile Motion Problem?

A shell leaves a mortar with a muzzle velocity of 500ft/s directed upward at 30degrees with the vertical. Determine the position of the shell and its resultant velocity 20s after firing. How high will it rise?

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  • 9 years ago
    Favorite Answer

    Final vertical position = Initial vertical velocity * time – ½ * g * time^2

    Final vertical velocity = Initial vertical velocity – g * time

    http://en.wikipedia.org/wiki/Standard_gravity

    According to the website above, the acceleration due to gravity = -32.174 ft/s^2

    As the mortar moves upward, it decelerates at 32.174 m/s^2.

    Initial vertical velocity = vi * cos θ = 500 * cos 30˚ = 433 ft/s

    At 20 seconds, vertical velocity = 433 – 32.174 * 20 = -210.48 ft/s

    Since the velocity is negative, the mortar has moved up to its highest position and is now moving downward at 210.5 ft/s.

    At 20 seconds, the vertical position = 433 * 20 – ½ * 32.174 * 20^2 = 6434.8 ft

    During the same time the horizontal velocity remains constant.

    Horizontal velocity = v * cos θ = 500 * sin 30˚ = 250 ft/s

    At 20 seconds, the horizontal position = 250 * 20 = 5000 ft

    At 20 seconds, the magnitude of the velocity = (-210.48^2 + 250^2)^0.5 = 326.8 ft/s

    The tangent of the angle from vertical = 250 ÷ 210.48

    Angle from vertical = 49.9˚

    At the maximum height, the vertical velocity is 0 ft/s

    vf = vi – g * t

    0 = 433 – 32.174 * t

    433 = 32.174 * t

    Time to reach maximum height = 13.46 seconds

    Final vertical position = Initial vertical velocity * time – ½ * g * time^2

    Maximum height = 433 * 13.46 – ½ * 32.174 * 13.46^2

    Maximum height = 2913.7 ft

  • Amy
    Lv 7
    9 years ago

    Use trigonometry to figure out the vertical and horizontal components of the initial velocity.

    Horizontal velocity remains constant.

    Calculate the horizontal position after 20s using the equation x = vt

    Vertical velocity is subject to a downward acceleration due to gravity.

    Calculate the vertical position after 20s using the equation y = (v0)t - 16 t^2

    (the -16 is the acceleration due to gravity in ft/s^2)

    Also calculate the vertical component of velocity after 20s. Combine this with the horizontal component to find the speed and direction.

  • 9 years ago

    Use Vf^2 = Vi^2 + 2*a*d where Vf = 0 at peak height, Vi = Vyi = 500sin60 = 433ft/s, a = -32ft/s^2 and d is the peak height

    d = 433^2/64 = 2930ft

    When does the mortar reach peak height?

    Vy(t) = Vyi - g*t = 0 at peak height => t = 433/32 = 13.5s

    Ignoring air resistance, the time to peak height is equal to the time from peak height back to launch height. Total flight time = 2*13.5 = 27s

    Vy(20) = 433- 32*20 = -207ft/s which means downward.

    Let height be Y(t)

    Y(t) = Yi + Vyi*t - 4.9*t^2 where Yi=0, Vyi = 500sin60, t = 20s

    Y(20) = 433*20 - 16*20^2 = 2260

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