Pergunta

2. A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.00m/s at an angle of 20.0^circ below the horizontal. It strikes the ground 3 .00 s later. (a) How far horizontally from the base of the building does the ball strike the ground? (b) Find the height from which the ball was thrown. (c)How long does it take the ball to reach a point 10.0 m below the level of launching?
Solução

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LucasProfissional · Tutor por 6 anos
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(a) To find the horizontal distance the ball travels before striking the ground, we can use the formula:<br /><br />Horizontal distance = (Initial velocity * cos(angle)) * Time<br /><br />Given:<br />Initial velocity = 8.00 m/s<br />Angle = 20.0°<br />Time = 3.00 s<br /><br />Plugging in the values, we get:<br /><br />Horizontal distance = (8.00 m/s * cos(20.0°)) * 3.00 s<br />Horizontal distance ≈ 23.1 m<br /><br />Therefore, the ball strikes the ground approximately 23.1 meters horizontally from the base of the building.<br /><br />(b) To find the height from which the ball was thrown, we can use the formula:<br /><br />Height = (Initial velocity * sin(angle)) * Time - 0.5 * g * Time^2<br /><br />Given:<br />Initial velocity = 8.00 m/s<br />Angle = 20.0°<br />Time = 3.00 s<br />g = 9.81 m/s^2 (acceleration due to gravity)<br /><br />Plugging in the values, we get:<br /><br />Height = (8.00 m/s * sin(20.0°)) * 3.00 s - 0.5 * 9.81 m/s^2 * (3.00 s)^2<br />Height ≈ 4.9 m<br /><br />Therefore, the height from which the ball was thrown is approximately 4.9 meters.<br /><br />(c) To find the time it takes for the ball to reach a point 10.0 m below the level of launching, we can use the formula:<br /><br />Time = sqrt(2 * Height / g)<br /><br />Given:<br />Height = 10.0 m<br />g = 9.81 m/s^2<br /><br />Plugging in the values, we get:<br /><br />Time = sqrt(2 * 10.0 m / 9.81 m/s^2)<br />Time ≈ 1.43 s<br /><br />Therefore, it takes approximately 1.43 seconds for the ball to reach a point 10.0 m below the level of launching.
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