kinematics

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A projectile is launched with an initial velocity of 25 m/s at a launch angle of 36
degrees. What is the magnitude of the projectile’s velocity when it is 7.5 m above the ground?

  • kinematics -

    v0=25 m/s
    θ=36° (assumed with horizontal)
    vh=v0 cos(θ)
    vv=v0 sin(θ)

    Let v=vertical velocity at any height S from ground, and g=9.81 m/s^2
    Use v²-vv²=-2gS (for vertical direction)
    v²=vv²-2gS
    By kinematics:

    Magnitude of Velocity,V, is the vector
    sum of the vertical and horizontal velocities, sqrt(v²+vh²).
    Note that vh is constant with time, and
    vv²+vh²=v0 (25 m/s)

    V=sqrt(v²+vh²)
    =sqrt(vv²-2gS+vh²)
    =sqrt(v0²-2gS)
    =sqrt(25²-2*9.81*7.5)
    =21.86 m/s
    Note that the magnitude does not depend of the angle θ.


    By energies:
    Initial kinetic energy = (1/2)m(v0²)
    Initial potential energy = 0 (assumed)
    Final potential energy = mgH (H=7.5m)
    Final Kinetic energy = (1/2)mv0²-mgH
    Final magnitude of velocity
    =sqrt(2(kinetic energy)/m)
    =sqrt(2((1/2)25^2-9.81*7.5))
    =sqrt(477.85)
    =21.859
    as before.

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