1) 45° is the ideal launch angle for projectile motion that gives the greatest horizontal range. Complementary angles will give the same range but will be less than the range for 45° Pick two complementary angles such that the sum of the angles is equal to 90°. Prove numerically the range is identical for both cases. Will the time aloft be the same for both angles? 2.) For the same two complementary angles assume the projectiles leave from a cliff and land 10.0m below the starting point. Prove the ranges in this case are not identical.

1) 45° is the ideal launch angle for projectile motion that gives the greatest horizontal range. Complementary angles will give the same range but will be less than the range for 45° Pick two complementary angles such that the sum of the angles is equal to 90°. Prove numerically the range is identical for both cases. Will the time aloft be the same for both angles? 2.) For the same two complementary angles assume the projectiles leave from a cliff and land 10.0m below the starting point. Prove the ranges in this case are not identical.

Name: Pls show your work 1) 45° is the ideal launch angle for projectile mot
Uploaded: 2024-03-20T06:57:47.000Z
Channel: Bartleby
Description: Pls show your work 1) 45° is the ideal launch angle for projectile motion that gives the greatest horizontalrange. Complementary angles will give the same range but will be less than the rangefor 45°●Pick two complementary angles such that the sum of

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)...

See similar textbooks

Similar questions

Boyles law inversely V =100cm P=30 Find the p when V is 75
centroid af about Find the shaded given axi. area
Needs Complete typed solution with 100 % accuracy.
If If A.B=0, What is the angle between Aand B O a. 90 degree O b. 120 degree O c. 45 degree O d. 0 degree
A bex welghis49 N16 pulk d 60mup along 037 ec plane Whats tew.kdone by the weight (gravlrational force) of the box? Select one O D.14x10
Convert 110mi/hr to ft/s. Show all of your work.
6.2-5 Refraction at the Brewster Angle. Use the condition n1 sec 01 = n2 sec 02 and Snell's law, n, sin 61 = n2 sin 62, to derive (6.2-12) for the Brewster angle. Also show that at the Brewster angle, 0 + 02 = 90°, so that the directions of the reflected and refracted waves are orthogonal, and hence the electric field of the refracted TM wave is parallel to the direction of the reflected wave. The reflection of light may be regarded as a scattering process in which the refracted wave acts as a source of radiation generating the reflected wave. At the Brewster angle, this source oscillates in a direction parallel to the direction of propagation of the reflected wave, so that radiation cannot occur and no TM light is reflected.
A. 590 N B. 1590 N C. 950 N