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.
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.
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd3392138-4176-488d-9d90-dfd88dbff0d2%2Fa7b79447-26e9-45c0-96d5-0f1d63ed23e0%2Fy1mhnew_processed.png&w=3840&q=75)
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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