A baseball is hit with a speed of 30m/s at an angle of 42degrees. It lands on the flat roof of a 14.0m tall nearby building. If the ball was hit when it was 1.4m above the ground, what horizontal distance does it travel before it lands on the building?

Question

anonymous · Answer

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verticle component of the velocity is usinX and horizontol component of the velocity is ucosX.
           then for vertical motion:
                                              h=usinx*t-1/2gt^2..... put x=42 degree,u=30m/sec ,g=10 and h=(14-1.4)=12.6m.and get the value of t.
 for horizontal motion the velocity is uniform because there is no acceleration in horizontal direction..
               hence R=ucosX*t and put the value of t,u and X in this equation and get the value of R which is horizontal distance travelled by the ball before landing.