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Q. 73

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Found in: Page 63

### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

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 $\frac{h}{d}g$

See the step by step solution

## Step 1. To write the given information,

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 $t$

## Step 2. To determine the acceleration of the ball

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

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

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)

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

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