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Students launch a steel ball horizontally from a tabletop. The initial horizontal speed of the ball is v, the tabletop height above the floor is H. The objective is to get steel ball into the coffee can of height h. Calculate the distance D at which the students need to place the can to make a "bulls-eye". v = 6 m/s H = 5 m h = 0.17 m Use g = 10 m/s2, enter two digits after decimal.

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Answer:

5.9 m

Explanation:

v=6m/s

H=5m

h=0.17m

g=10 m/s

First we find the time in which ball reaches the entrance of coffee can. For this purpose the ball has to travel a distance  S vertically downward such that

S= 5-0.17 = 4.83 m

While falling down Vi =0

g= 10 m/sec2

t = ?

S=vit+1/2 gt2

==> 4.83=0+0.5×10×[tex]t^{2}[/tex]

==> 4.83=5×[tex]t^{2}[/tex]

==> [tex]t^{2}[/tex] =4.83/5 = 0.97

==> t= 0.98 sec

So in this time the ball reaches the coffee can height level. Now let's calculate horizontal distance covered by the ball in this time.

Horizontal distance = horizantal velocity × time

= 6 × 0.98 = 5.9 m

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