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physics

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A man can throw a ball a maximum horizontal
distance of 97.8 m.
The acceleration of gravity is 9.8 m/s2 .
How far can he throw the same ball verti-
cally upward with the same initial speed?
Answer in units of m.

  • physics - ,

    The maximum horizontal distance is obtained by throwing the ball at 45° with the horizontal.
    Let v0=initial speed,
    resolve v0 into horizontal and vertical components,
    vh=v0*cos(45)
    vv=v0*sin(45)

    The time, t, the ball was in the air can be calculated by the horizontal distance divided by the horizontal component of velocity, vh.
    t=97.8/vh

    The vertical distance can be calculated using
    S=vv*t+(1/2)gt²
    where g=-9.8 m/s² (downwards)
    Substitute vv and t into the above equation gives
    S=(v0 sin45)*(97.8/(v0 cos45) + (1/2)g (97.8/(v0 cos45))²
    From which v0 can be calculated as approximately 31 m/s.

    Now apply the same velocity in the vertical direction,
    initial kinetic energy
    KE = (1/2)mv0²
    At the top, h metres above ground, kinetic energy is zero, but potential energy is
    PE = mgh
    Equate KE=PE,
    mgh=(1/2)mv0²
    h=(1/2)mv0²/(mg)
    =(1/2)v0²/g
    = 49 m, approx.

  • physics - ,

    First you need to know what angle from horizontal gives maximum range. With no air friction it is 45 degrees but you need to show that.
    V = throwing speed
    u = V cos A
    Vo = V sin A
    R = range = u T = V T cos A = 97.8
    speed up = Vo - (9.8) t
    = 0 at top where t = T/2 because half the time is rising and half falling
    so
    0 = Vo - 4.9 T
    V sin A = 4.9 T
    so T = .204 V sin A
    range = V T cos A
    range = .204 V^2 sin A cos A
    so for what angle A is sin A cos A a maximum?
    y = sin x cos x
    dy/dx = 0 at max = -sin^2 x + cos^2 x
    that is where the sin is equal to the cosine, 45 degrees, where sin = cos = .707 and sin*cos = .707*.707 = .5
    so now we do the problem
    range = .204 V^2 (.5) = 97.8
    V^2 = 959
    31 m/s approx
    NOW throw vertical up at 31 m/s
    31 = 9.8 t at top
    t = 3.16 seconds traveling up
    h = V t - 4.9 t^2
    = 31*3.16 - 4.9(3.16^2)
    = 97-49 = 48 meters approximately

  • physics - ,

    MathMate's energy method is far cooler than my more basic method.

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