# Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems: 37

The ball reaches a height of 3.08 meters above the top of the tube.

#### Work Step by Step

We can find the acceleration of the ball while it is in the tube; $\sum F = ma$ $F_{air}- mg = ma$ $a = \frac{F_{air}- mg}{m}$ $a = \frac{2.0~N- (0.050~kg)(9.80~m/s^2)}{0.050~kg}$ $a = 30.2~m/s^2$ We can find the speed as the ball leaves the tube; $v^2 = 2ay$ $v = \sqrt{2ay}$ $v = \sqrt{(2)(30.2~m/s^2)(1.0~m)}$ $v = 7.77~m/s$ We can find the maximum height the ball reaches above the tube. We can let $v_0 = 7.77~m/s$ for this part of the question. $y = \frac{v^2-v_0^2}{2g}$ $y = \frac{0-(7.77~m/s)^2}{(2)(-9.80~m/s^2)}$ $y = 3.08~m$ The ball reaches a height of 3.08 meters above the top of the tube.

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