A physics book slides off a horizontal table top with a speed of 1.25m/s . It st

Question

A physics book slides off a horizontal table top with a speed of 1.25m/s . It strikes the floor after a time of 0.340s . Ignore air resistance

1-Find the height of the table top above the floor.
2-Find the horizontal distance from the edge of the table to the point where the book strikes the floor.
3-Find the horizontal component of the book's velocity just before the book reaches the floor.
4-Find the vertical component of the book's velocity just before the book reaches the floor.
5-Find the magnitude of the book's velocity just before the book reaches the floor.
6-Find the direction of the book's velocity just before the book reaches the floor.

Explanation / Answer

1- The time it takes to fall only depends on it's height and vertical speed. Since it had purly horizontal velocity; the vertical velocity was zero. D = vot + 1/2 a t^2 vo vert = 0 D = 1/2*9.81m/s^2*(0.340s)^s = 0.567m height of the table top above the floor 0.567m 2- sideways speed does not change (gravity is perpendicular to sideways motion) so speed is constant. distance = rate times time d = 1.25m/s*0.340s = 0.425m horizontal distance is 0.425m 3- this never changes, so it is still 1.25m/s 4- want to find final velocity Vf = Vo + at = 0 + 9.81m/s^2*0.340s = 3.34m/s down 5-add the two perpendicular components using Pythagorian's Theorem. Vtot = SQRT(Vf vert^2 + V hor^2) = SQRT(3.34^2 + 1.25^2) m/s = 3.56m/s 6- hor = 1.25 vert = 3.34 book is going more down than sideways, so angle will be more than 45 deg below horizontal. tan(theta) = 3.34/1.25 theta = INV TAN(3.34/1.25) = 69.4 degrees down from horizontal

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