Physics. Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 39.0° below the horizontal. If it str?

Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 39.0° below the horizontal. If it strikes the ground 28.1 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)

(a) the time of flight

(b) the initial speed

(c) the speed and angle of the velocity vector with respect to the horizontal at impact

the height from the other problem was h=41.0 m

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  • Anonymous
    2 months ago

    Vox*t = 28

    t = 28/Vox

    -41 = -Voy*t-g/2*t^2

    41 = Voy*28/Vox+4.9*28^2/Vox^2

    Voy/Vox = tan 39°= 0.810

    41-28*0.810 = 4.9*28^2/Vox^2

    Vox = √4.9*28^2 / (41-28*0.810) = 14.48 m/sec

    initial speed Vo = Vox/cos 39 = 14.48/0.777 = 18.63 m/sec

    time of flight t = d/Vox = 28/14.48 = 1.934 sec

    final vert. speed Vy = -Voy-g*t = -18.63*sin 39-9.8*1.934 = -37.0 m/sec

    impact speed V = √Vy^2+Vox^2 = 39.7 m/sec

    impact angle Θi = arctan (Vy/Vox) = arctan(-37/14.48) = -68.6°

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  • 2 months ago

    It might be a good idea to do what they suggested

    y = h - v0t(sin39°) - 0.5gt^2

    Take the horizontal range X which seems to be 28.1 m (even though it clearly sounds like the hypotenuse is 28.1 m)

    v0t(cos39°) = X

    0 = h - Xtan39° - 0.5gt^2

    t^2 = 2(h - Xtan39°)/g

    v0 = X / (t cos39°)

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