A physics book slides off a horizontal tabletop with a speed of 2.70 m/s . It st

ID: 1594648 • Letter: A

Question

A physics book slides off a horizontal tabletop with a speed of 2.70 m/s . It strikes the floor in 0.350 s. Ignore air resistance.

Part A

Find the height of the tabletop above the floor. h=

Part B

Find the horizontal distance from the edge of the table to the point where the book strikes the floor. L=

Part C

Find the horizontal component of the book's velocity, just before the book reaches the floor. vx =

Part D

Find the vertical component of the book's velocity just before the book reaches the floor. vy=

Part E

Find the magnitude of book's velocity, just before the book reaches the floor. v=

Part F

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

initial speed of the book = 2.7 m/s

using the equation of linear motion

Y = Uy*t + 0.5*(-g)t^2

y = 0 + 0.5*(-9.8)*(0.35)^2

y = -0.6m

height of the tabletop above the floor = 0.6 m

B) X = Uxt +0.5ax*t^2

but acceleration in x- direction = 0

x = 2.7*0.35 = 0.945 m

c) horizontal component of books velocity before reaching the floor

Vx = Ux = 2.7 m/s

d) vertical component of the book's velocity just before the book reaches the floor.

Vy = Uy - gt

Vy =0 - 9.8*0.35 = -3.43 m/s

e) magnitude of books velocity = sqrt( 2.7^2 + 3.43^2) = 4.365 m/s

f) angle =tan^-1 ( 3.43/2.7) = 51.8 degree below the horizontal

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.