Students in a lab measure the speed of a steel ball launched horizontally from a

Question

Students in a lab measure the speed of a steel ball launched horizontally from a tabletop to be (v). The table-top is distance (y) above the floor. They place a tall tin coffee can of height (0.1 y)on the floor to catch the ball.

Part A

Find a horizontal distance from the base of the table the can should be placed.

Express your answer in terms of the variables ( exttip{v}{v}), ( exttip{y}{y}), and appropriate constants.

Part B

If the ball leaves the tabletop at a speed of 4.8 ({ m m/s}) , the tabletop is 1.6 ({ m m}) above the floor, and the can is 0.15 ({ m m}) tall, find a horizontal distance from the base of the table that the center of the can should be placed.

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

along vertical

inital velocity voy = 0

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

displacement dy = yf - yi

from equation of motion

dy = voy*t + 0.5*ay*t^2

t = sqrt(2*dy/ay)

along horizantal

X = vox*t

x = v*sqrt(2*(0.1y-y)/ay

x = v*sqrt(2*(y-0.1y)/g)

++++++++++++++++++++

part B

x = 4.8*sqrt(2*(1.6-0.15)/9.8)

x = 2.61 m <<------------answer

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