A physics book slides off a horizontal tabletop with a speed of 2.30 m/s. It str

Question

A physics book slides off a horizontal tabletop with a speed of 2.30 m/s. It strikes the floor in 0.350 s. Ignore air resistance.

Part A
Find the height of the tabletop above the floor.

Part B
Find the horizontal distance from the edge of the table to the point where the book strikes the floor.

Part C
Find the horizontal component of the book's velocity, just before the book reaches the floor.

Part D
Find the vertical component of the book's velocity just before the book reaches the floor.

Part E
Find the magnitude of book's velocity, just before the book reaches the floor.

Part F
Find the direction of the book's velocity, just before the book reaches the floor.

Explanation / Answer

a) irrespective of the horizontal speed, the inital veritcal speedis zero. time taken to reach is 0.350 seconds. so S = ut +0.5at2 = 0.5 * 9.81 * 0.350 * 0.350 = 0.6 meters (ht oftable) b) Since the horizontal velocity is uniform always,multiplying it with time gives the horizontal distance travelled tothe point where it strikes the floor = 1.1 * 0.350 = 0.385meters c) There is no change in horizontal component. It remains samealways 1.1 m/s d) Vertical component of velocity is found out by treating thebook as a object falling vertically down under the influence ofgravity and using V = u + at = 0 + 9.81 * 0.350 = 3.43 m/s e) Using conservation of energy principle, energy at top =energy at bottom PE top + KE top = KE Bottom 9.81*0.6 + 0.5*1.1*1.1 = 0.5*V*V (as mass is equal, itgetscancelled) V = 3.603 m/s Or simply use V = sq root(Horizontal comp sq + veritcal compsq) f) 45 degrees below the horizontal (but plz check this one. Iam not sure

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