Post a New Question

physics

posted by .

Suppose that a shot putter can put a shot at the worldclass speed v0 = 14.00 m/s and at a height of 2.160 m. What horizontal distance would the shot travel if the launch angle θ0 is (a) 45.00° and (b) 42.00°? The answers indicate that the angle of 45°, which maximizes the range of projectile motion, does not maximize the horizontal distance when the launch and landing are at different heights.

  • physics -

    a. Vo = 14m/s[45o]

    Xo = 14*cos45 = 9.9 m/s.
    Yo = 14*sin45 = 9.9 m/s.

    Y = Yo + g*t = 0
    Tr=Yo/g = 9.9/9.8 = 1.01 s. = Rise time.

    hmax = ho + Yo*t + 0.5g*t^2
    hmax=2.16 + 9.9*1.01 - 4.9*1.01^2=7.16m
    Above gnd.

    h = 0.5g*t^2 = 7.16
    4.9t^2 = 7.16
    t^2 = 1.46
    Tf = 1.21 s. = Fall time.

    D = Xo * (Tr+Tf)=9.9 * (1.01+1.21=21.9m
    = Hor. distance.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question