How is he getting 1.21 m/s Figure 4 for a diagram of the projectile motion.Themaximumheight ofthelaunchbecalculatedcangiven thatvy2 = 0 m/s,vyl = 8.71sin81.21 m/s,dy = 0.491 m anda = -9.8 m/s2 -usingtheformulahmax = (vy22 - vyl)/2a) + dy . After calculations, the maximum height of the projectile is 0.566m.Finally, the impact velocity can be calculated by first finding the vy2 using vy2 = Vvy1 + 2ad - given thatvyl =-8.71sin8 =- 1.21m/s, a =-9.8 m/s2 and d =-0.491m. After calculations, vy2 = 3.33 m/s. To find theimpact velocity, the horizontal and vertical components of velocity mustbe added [V impact = Vvx² +vy2 ] to give Vimpact = 9.25 m/s @21 ah. Refer to Figure 4 for a diagram of theprojectile motion.op

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How is he getting 1.21 m/s

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

How is he getting 1.21 m/s

Figure 4 for a diagram of the projectile motion.
The
maximum
height of
the
launch
be
calculated
can
given that
vy2 = 0 m/s,
vyl = 8.71sin8
1.21 m/s,
dy = 0.491 m and
a = -9.8 m/s2 -
using
the
formula
hmax = (vy22 - vyl)/2a) + dy . After calculations, the maximum height of the projectile is 0.566m.
Finally, the impact velocity can be calculated by first finding the vy2 using vy2 = Vvy1 + 2ad - given that
vyl =-8.71sin8 =- 1.21m/s, a =-9.8 m/s2 and d =-0.491m. After calculations, vy2 = 3.33 m/s. To find the
impact velocity, the horizontal and vertical components of velocity must
be added [
V impact = Vvx² +vy2 ] to give Vimpact = 9.25 m/s @21 ah. Refer to Figure 4 for a diagram of the
projectile motion.
op

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