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Calculate the net force acting on the rocket during the launch process if the mass of the rocket is 25 g and the approximate launch time is 0.14 seconds.  C. Explain how you could calculate the launch force acting on the rocket but what would you have to ignore and why? DATA FOR B IS IN IMAGE

Calculate the net force acting on the rocket during the launch process if the mass of the rocket is 25 g and the approximate launch time is 0.14 seconds.  C. Explain how you could calculate the launch force acting on the rocket but what would you have to ignore and why? DATA FOR B IS IN IMAGE

University Physics Volume 1
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter13: Gravitation
Section: Chapter Questions
Problem 85CP: Show that for small changes in height h, such that hRE , Equation 13.4 reduces to the expression...
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B. Calculate the net force acting on the rocket during the launch process if the mass of the rocket is 25 g and the approximate launch time is 0.14 seconds. 

C. Explain how you could calculate the launch force acting on the rocket but what would you have to ignore and why?

DATA FOR B IS IN IMAGE

 

Using conservation of energy, К. Е., +Р. Е.; — К. Е.f +P. E.f mv + mgh, = mv + mgh, where, v; = launch speed maximum height h; = initial height = 0 m hf %3| velocity at maximum height = 0m ·s- Vf = ( At maximum height velocity of projectile is 0. substituting values, zmv + mg (0 m) = ;m (0 ms-') + mgh; So, above reduces to, v = gh; So, Iaunching velocity (v;) = V2 gh; substituting numerical values, Iaunching velocity (v;) = \/2 (9. 8 m · s-2 ) (1. 47 m) = 5. 37 m · s-1
Transcribed Image Text:Using conservation of energy, К. Е., +Р. Е.; — К. Е.f +P. E.f mv + mgh, = mv + mgh, where, v; = launch speed maximum height h; = initial height = 0 m hf %3| velocity at maximum height = 0m ·s- Vf = ( At maximum height velocity of projectile is 0. substituting values, zmv + mg (0 m) = ;m (0 ms-') + mgh; So, above reduces to, v = gh; So, Iaunching velocity (v;) = V2 gh; substituting numerical values, Iaunching velocity (v;) = \/2 (9. 8 m · s-2 ) (1. 47 m) = 5. 37 m · s-1
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