Physics

A ball player hits a home run, and the baseball just clears a wall 17.0 m high located 124.0 m from home plate. The ball is hit an an angle of 35° to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1.0 m above the ground.

(a) What is the initial speed?
m/s
(b) How much time does it take for the ball to reach the wall?
s
(c) Find the components of the velocity and the speed of the ball when it reaches the wall
vy,f = m/s
vx,f = m/s
vf = m/s

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  1. a. Dh=Vo^2*sin(2A) / g = 124m = Hor. dist.
    Vo^2^sin(79) / 9.8 = 124,
    0.9397Vo^2 / 9.8 = 124,
    0.9397Vo^2 = 1215.2,
    Vo^2 = 1293.2,
    Vo = 36m/s @ 35 = Initial velocity.
    Xo = 36cos35 = 29.49m/s.
    Yo = ver. = 36sin35 = 20.65m/s

    c. hmax = (Yf^2 - Yo^2) = 2g,
    hmax = (0 - (20.65)^2) / -19.6 = 21.8m.

    Yf^2 = Yo^2 + 2g*d,
    Yf^2= 0 + 19.6*(21.8-17)=94.1,
    Yf = 9.7m/s. = Vy,f.

    Xo = 29.49m/s = Vx,f.

    Vf = sqrt((29.49)^2 + (9.7)^2) = 31m/s.

    b. t(up) = (Yf - Yo) / g,
    t(up) = (0 - 20.65) / -9.8 = 2.1071s.

    t(dn) = = (Yf - Yo) / g,
    t(dn) = (9.7 0) / 9.8 = 0.9898s.

    T = t(up) + t(dn)
    T = 2.1071 + 0.9898 = 3.1s to reach te
    wall.

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    posted by Henry

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