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A brick is thrown upward from the top of a building at an angle of $25^{\circ}$ to the horizontal and with an initial speed of 15 $\mathrm{m} / \mathrm{s}$ . If the brick is in flight for $3.0 \mathrm{s},$ how tall is the building?

25$\mathrm { m }$

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Cornell University

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

McMaster University

So we're going to of course, choose our origin to be the initial position of the projectile. And after three seconds, it is at ground level. So we can say the vertical displacement delta Y is going to be equaling negative H because it falls down a certain height h to find H, we'll use Delta y equaling fee. Why initial t plus 1/2 What G t squared now here. We can then say that negative h will equal. This will be 15 meters per second, multiplied by sine of 25 degrees. This would be multiplied by 3.0 seconds, plus 1/2 multiplied by negative 9.80 meters per second squared multiplied by 3.0 seconds. Quantity squared and we find that then H is equaling 25 meters. So the projectile falls a distance of 25 meters. That is the end of the solution. Thank you for watching