Question

A physics book slides off a horizontal table top with a speed of 1.30 m/sm/s. It strikes the floor after a time of 0.410 ss. Ignore air resistance.

Find the height of the table top above the floor.

Express your answer with the appropriate units.

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

Express your answer with the appropriate units.

Find the magnitude of the horizontal component of the book's velocity just before the book reaches the floor.

Express your answer with the appropriate units.

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

Express your answer with the appropriate units.

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

Express your answer with the appropriate units.

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

Express your answer with the appropriate units.

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

Answer 1

First thing, the book slides off horizontally, i.e. u_{​​​​​​x}= 1.3m/s.

And U_{​​​​​​y} = 0m/s

Now solving the question, using 1-D kinematics in horizontal as well as vertical direction,

Height: we use 2nd equation of motion in vertical direction.

As the data is given that it reaches floor in 0.410 s

So, t = 0.410s

U_{​​​​​​y}=0, & a_{​​​​​​y}= g=10m/s^{​​​​​​2}

H = U_{​​​​​​y}*t + (0.5)*g*t^{​​​​​​2}

H = (0.5)*10*(0.41)² = 0.84m

Horizontal distance: we use 2nd equation of motion in horizontal direction,

U_{​​​​​​x} = 1.30

t = 0.410s

a_{​​​​​​x} = 0

Therefore,

S = U_{​​​​​​x}*t + (0.5)*a_{​​​​​​x}*t^{​​​​​​2}

S = 1.3*0.41 = 0.533m

Vertical velocity before reaching : for this we use 1st equation of motion,

V_{​​​​​​y} = U_{​​​​​​y} + a_{​​​​​​y}*t

V_{​​​​​​y} = g*t = 10*0.41 = 4.1m/s

Horizontal velocity before reaching : again we use first equation of motion

V_{​​​​​​x} = U_{​​​​​​x} + a_{​​​​​​x}*t

V_{​​​​​​x} = U_{​​​​​​x} = 1.3m/s

Magnitude of net velocity:. We use formula or resultant:

When two vectors having an angle of 90° are added their magnitude becomes:

V= √{(V_{​​​​x}​)² + (V_{​​​​​​y})²}

V = √{(1.3)² + (4.1)²} = 4.30m/s

Direction: it is found by vector property

The direction of velocity vector, is below the horizontal by an angle given by:

= tan^{​​​​-1}(V_{​​​​​​y}/V_{​​​​​x}) = tan^-1(4.1/1.3) = 72.4° below the horizontal.