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A ball is launched from the base of a plane with inclination ϴ = 45º Determine (a) acceleration on ascent and on frictionless descent. (b) acceleration in ascent and descent with friction (c) If the ball is thrown with a velocity V0 = 20 m/s determine the rise times with and without friction. likely answers to letter a and b I just wanted the answer to the letter C likely answers to letter a and b (image)

A ball is launched from the base of a plane with inclination ϴ = 45º Determine (a) acceleration on ascent and on frictionless descent. (b) acceleration in ascent and descent with friction (c) If the ball is thrown with a velocity V0 = 20 m/s determine the rise times with and without friction. likely answers to letter a and b I just wanted the answer to the letter C likely answers to letter a and b (image)

Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter3: Motion In Two Dimensions
Section: Chapter Questions
Problem 5OQ: Does a car moving around a circular track with constant speed have (a) zero acceleration, (b) an...
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A ball is launched from the base of a plane with inclination ϴ = 45º

Determine

(a) acceleration on ascent and on frictionless descent.

(b) acceleration in ascent and descent with friction

(c) If the ball is thrown with a velocity

V0 = 20 m/s determine the rise times with and without friction.

likely answers to letter a and b

I just wanted the answer to the letter C

likely answers to letter a and b (image)

mg sin 0 + man = 0 where, as is the ascent acceleration, hence, aa = -g sin 0 aA = -g sin 45° negative sign as the body is decelerating as it goes against gravity Now the decent acceleration, from FBD (b) mg sin 0 — тар —D 0 where, ap is the decent acceleration, hence, ap =g sin 0 ap = g sin 45° ap
Transcribed Image Text:mg sin 0 + man = 0 where, as is the ascent acceleration, hence, aa = -g sin 0 aA = -g sin 45° negative sign as the body is decelerating as it goes against gravity Now the decent acceleration, from FBD (b) mg sin 0 — тар —D 0 where, ap is the decent acceleration, hence, ap =g sin 0 ap = g sin 45° ap
f = µN wnere, N = mg cos 0 (Normal force) hence, f = u mg cos 0 now, mg sin 0 +f + man = 0 where, a, is the ascent acceleration, hence, an = - (ug cos 0 + g sin Ø) negative sign as the body is decelerating as it goes against gravity an = ug cos 45° + g sin 45° (u + 1) For the acceleration of the decent, mg sin 0 =f + man where, ap is the decent acceleration, hence, map = mg sin 0 - f map = mg sin 0 - umg cos 0 ap = -ug Cos 0 + g sin 0 ap = -ug Cos 45° + g sin 45° (1 – ) up =
Transcribed Image Text:f = µN wnere, N = mg cos 0 (Normal force) hence, f = u mg cos 0 now, mg sin 0 +f + man = 0 where, a, is the ascent acceleration, hence, an = - (ug cos 0 + g sin Ø) negative sign as the body is decelerating as it goes against gravity an = ug cos 45° + g sin 45° (u + 1) For the acceleration of the decent, mg sin 0 =f + man where, ap is the decent acceleration, hence, map = mg sin 0 - f map = mg sin 0 - umg cos 0 ap = -ug Cos 0 + g sin 0 ap = -ug Cos 45° + g sin 45° (1 – ) up =
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