Question

A physics book slides off a horizontal tabletop with a speed of 1.60 m/s. It strikes the floor in 0.350 s. Ignore air resistance. Find:

a. The height of the tabletop above the floor.
b. The horizontal distance from the edge of the table to the point where the books strikes the floor.
c. The horizontal and vertical components of the book.

Answer 1

Answer:

Explanation:

Given

Initial horizontal speed of book $u_x=1.6\ m/s$

time taken to reach $t=0.35\ s$

Considering vertical motion

$h=ut+\frac{1}{2}at^2$

h=displacement

u=initial velocity

a=acceleration

t=time

$h=0+\frac{1}{2}\times 9.8\times (0.35)^2$

$h=0.6\ m$

(b)horizontal distance moved

$x=u_x\times t$

$x=1.6\times 0.35$

$x=0.56\ m$

(c)Horizontal component of velocity will remain same as there is no acceleration in horizontal direction

vertical velocity at bottom is

$v_y=0+gt$

$v_y=9.8\times 0.35=3.43\ m/s$

$v_x=u_x=1.6\ m/s$

$v_{net}=\sqrt{v_x^2+v_y^2}$

$v_{net}=\sqrt{1.6^2+3.43^2}$

$v_{net}=\sqrt{14.325}$

$v_{net}=3.78\ m/s$