a. using the launch velocity determined in question 1 or 2 (should be the same!) determine the maximum range of the rocket, with a launch angle of 45 degrees. *Assume level ground.b. Compare the value in question 4 with the value for maximum range in the table. Discuss reasons for differences. (even if they happen to be exactly the same!) QUESTION ONE DATA IN PICTURE. Using conservation of energy,К. Е., +Р. Е.; — К. Е.f +P. E.fmv + mgh, = mv + mgh,where, v; = launch speedmaximum heighth; = initial height = 0 mhf%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

The launching velocity v1 = 5.37 m.s-1 The angle of rocket launching θ=45°

Answered: using the launch velocity determined in…

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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a. using the launch velocity determined in question 1 or 2 (should be the same!) determine the maximum range of the rocket, with a launch angle of 45 degrees. *Assume level ground.

b. Compare the value in question 4 with the value for maximum range in the table. Discuss reasons for differences. (even if they happen to be exactly the same!)

QUESTION ONE DATA IN PICTURE.

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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