Question

A book slides off of the top of a counter 1.5 meters above the ground and lands in a moving cart with a velocity of 2.0 meters per second in the positive x direction. The cart started at t=0 s directly below the table edge.

What is the final x position where the book lands in the cart?
The cart has a height of 20 centimeters. What time does the book
land in the cart?

What is the final velocity in polar form of the book when it lands
in the cart?

Answer #1

For the book

along vertical

initial velocity voy = 0

initial position yo = 1.5 meters

final position y = 20 cm = 0.2 m

acceleration ay = -g = -9.8 m/s^2

from equation of motion

y - yo = voy*t + (1/2)*ay*t^2

0.2 - 1.5 = 0 - (1/2)*9.8*t^2

time t = 0.515 seconds

the final x position where the book lands in the cart is v*t =
2*0.515 = 1.03 meters

time t = 0.515 seconds

--------------------------

along vertical

vy = voy + ay*t

vy = 0 - 9.8*0.515 = -5.047 m/s

vx = vo + ax*t = vox = 2 m/s

final velocity v = sqrt(5.047^2 + 2^2) = 5.43 m/s

angle theta = arctan(-5.047/2) = -68.4^{o}

v = ( 5.43 m/s , -68.4 ^{o} )

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