An object is dropped trom a tower, 1600 Htab the object's volocity at the moment of impact? It takes \( \square \) sec for the object to hit the groun (Round to the nearest tenth as needed.)
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When an object is dropped from a height, we can use the equation \( v = g \cdot t \), where \( v \) is the final velocity, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, m/s^2 \)), and \( t \) is the time taken to fall. To find the time it takes to hit the ground, we can use the formula \( h = \frac{1}{2} g t^2 \). Rearranging it for \( t \) gives \( t = \sqrt{\frac{2h}{g}} \). Plugging in the values, we get \( t = \sqrt{\frac{2 \cdot 1600}{9.81}} \), which gives you around 18.0 seconds for the fall. Therefore, the velocity at impact can be calculated as \( v = 9.81 \cdot 18.0 \), resulting in approximately 176.0 m/s just before hitting the ground. Keep in mind that this model assumes no air resistance, so in real-life situations, the object might fall slightly slower than calculated due to drag!