A ball is shot from the top of a building with an initial velocity of 19 m/s at

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Let: Vix => x-component of the initial velocity Viy => y-component of the initial velocity Vi => initial velocity t => time of flight ? => angle above the horizontal g => acceleration due to gravity, which is 9.8 m/s² Sx => range/horizontal distance Sy => height of the building a.) Vix and Viy = ? Vix = Vicos? Vix = 18cos42° Vix = 13.38 m/s Viy = Visin? Viy = 18sin42° Viy = 12.04 m/s b.) Sy = ? But we know that the horizontal distance between those buildings are 55 meters apart and the buildings have the same height. Before we determine the height of the building, we must first solve for the time of flight. Sx = Vix*t 55 = 13.38*t 4.11 secs = t Now we can solve for the height of the building. Sy = Viy*t + 0.5gt² Sy = 12.04(4.11) + 0.5(-9.8)(4.11)² Sy = -33.29 meters Reject the negative since because it indicates the displacement of the ball. Sy = 33.29 meters above the ground.

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