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A ball is shot from the top of a building with an initial velocity of 19m/s at an angle theta = 45 above the horizontal.

Vxo and Vyo, are 13.4m/s

If a nearby building is the same height and 50m away, how far below the top of the building will the ball strike the nearby building?

I found the time from x=vt, 50=13.4*t

t= 3.73s

then y= yo + Vyo*t - 0.5*at^2

y= yo + 13.4*3.73 - 0.5*9.8*(3.73)^2..

shouldn' t the answer with that part of formula which is " Vyo*t - 0.5*at^2 " give me how far it could be?

  • Physics - ,

    L=vₒ²•sin2α/g =19²sin2•45/9.8=36.8 m
    At the distance of 36.8 m from the first building the ball is at the level of starting point and has the velocity v(x)=v₀•cos45° =13.4 m/s
    The distance to the second building is 50-36.8 = 13.2 m
    this distance is covered for t=l/v(x) = 13.2/13.4 =0.99 s.
    h= gt²/2=9.8•0.99²/2=4.8 m.

  • Physics - ,

    ty :)

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