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Q. 73

Expert-verifiedFound in: Page 63

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

Your goal in the laboratory is to launch a ball of mass m straight up so that it reaches exactly the height h above the top of the launching tube. You and your lab partners will earn fewer points if the ball goes too high or too low. The launch tube uses compressed air to accelerate the ball over a distance d, and you have a table of data telling you how to set the air compressor to achieve the desired acceleration. Find an expression for the acceleration that will earn you maximum points

The acceleration of the ball to earn maximum points is $\frac{h}{d}g$

The maximum height the ball should reach is $h$The distance that the launching tube can accelerate is $d$Since the ball is thrown upwards, the final velocity is ${v}_{f}=0m/s$Let the initial velocity is ${v}_{i}$The acceleration of the ball is $a$The acceleration due to gravity is $g$The time taken by the ball to reach height is $t$

Write the equation of motion for the ball thrown above for the maximum height h,

$h=\frac{1}{2}g{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {t}^{2}=\frac{2h}{g}.....\left(1\right)$

Write the equation of motion to determine the initial velocity of the ball,

${v}_{f}={v}_{i}+at\phantom{\rule{0ex}{0ex}}0={v}_{i}-gt\phantom{\rule{0ex}{0ex}}{v}_{i}=gt......\left(2\right)$

Write the equation of motion to determine the acceleration,

localid="1648616761150" ${\left({v}_{f}\right)}^{2}-{\left({v}_{i}\right)}^{2}=2ad\phantom{\rule{0ex}{0ex}}0-{\left({v}_{i}\right)}^{2}=2ad......\left(3\right)$

Plug the value ${v}_{i}$ from equation (2)${\left(gt\right)}^{2}=2ad\phantom{\rule{0ex}{0ex}}{g}^{2}{t}^{2}=2ad......\left(4\right)$Substitute the value of ${t}^{2}$ form equation (1)

${g}^{2}\left[\frac{2h}{g}\right]=2ad\phantom{\rule{0ex}{0ex}}2gh=2ad\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{h}{d}g$

Therefore the acceleration of the ball to earn maximum points is $\frac{h}{d}g$

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