Post a New Question


posted by .

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,

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

    The vertical distance can be calculated using
    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,
    = 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
    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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question