ID 34568048enrolmentio - 30 ao Secure https://www.flipitphysics.com Course ViewP
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ID 34568048enrolmentio - 30 ao Secure https://www.flipitphysics.com Course ViewProblem?uniti te rorozca Tuesday, January 30 | 7:06 PM it macmillan learning Pima Community College Unit 3: Prelecture/Bridge / Problems Problems: Straight-Line Motion With Constant Acceleration Deadline: 100% until Tuesday, January 30 at 1:00 PM Problem 23 Alex climbs to the top of a tall tree while his friend Gary waits on the ground below. Alex throws down a ball with an initial speed of 4.10 m/s from 50.0 m above the ground. The starting height of the ball thrown upward is 1.50 m above the ground. ignore the effects of air resistance. 1) At what speed must Gary throw a ball up in order for the two balls to cross paths 25.0 m above the ground? (Express your answer to three significant figures.) m/s Submit You currently have O submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question.Explanation / Answer
Ball thrown down by Alex
Vi = 4.10 m/sec
a = g = 9.81 m/sec
when balls cross at 25 m, So vertical distance traveled = 50 - 25 = 25 m
Using equation
Vf^2 = Vi^2 + 2*a*s
Vf = sqrt (4.1^2 + 2*9.81*25) = 22.52 m/sec
Time taken to reach this speed will be
Vf = Vi + a*t
t = (22.52 - 4.1)/9.81 = 1.88 sec
Now when Garry throws ball upwards
Vi = ?
t = 1.88 sec
a = -9.81 m/sec
h = vertical distance traveled when balls cross = 25 - 1.5 = 23.5 m
Using equation
h = Vi*t + 0.5*a*t^2
23.5 = Vi*1.88 - 0.5*9.81*1.88^2
Vi = (23.5 + 0.5*9.81*1.88^2)/1.88
Vi = 21.7 m/sec