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celeration due to gravity near the surface of the earth can be considered constant, with a value of 9.80 m/s² downwards (towards the ground). a. b. d

celeration due to gravity near the surface of the earth can be considered constant, with a value of 9.80 m/s² downwards (towards the ground). a. b. d

College Physics
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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Acceleration due to gravity near the surface of the earth can be considered constant, with a value of 9.80 m/s² downwards (towards the ground). a. b.
Transcribed Image Text:Acceleration due to gravity near the surface of the earth can be considered constant, with a value of 9.80 m/s² downwards (towards the ground). a. b.
Part A If a ball is dropped from a height of d =29.0 m (as illustrated in "a," above) how long will i take for the ball to hit the ground? t = S Submit Previous Answers Not completed before the time limit Part B Suppose that instead of being dropped, the ball is thrown downwards (as illustrated in "b." above) from the same height d =29.0 m ) and the ball hits the ground 1.46 s after it is thrown. What was the initial velocity of the ball in this case? ΑΣφ ? Vo = m/s
Transcribed Image Text:Part A If a ball is dropped from a height of d =29.0 m (as illustrated in "a," above) how long will i take for the ball to hit the ground? t = S Submit Previous Answers Not completed before the time limit Part B Suppose that instead of being dropped, the ball is thrown downwards (as illustrated in "b." above) from the same height d =29.0 m ) and the ball hits the ground 1.46 s after it is thrown. What was the initial velocity of the ball in this case? ΑΣφ ? Vo = m/s
Expert Solution
Step 1

Part 1

Ball drop from height d=29m

Acceleration due to gravity= 9.8m/s2

Initial velocity u= 0m/s

From equation of motion 

v2= u2 + 2as

v2= 2×9.8×29

v= 23.84m/s

v= velocity when ball reach the ground

Now 

V= u+gt

23.84= 0+9.8×t

t= 23.84/9.8= 2.43 sec

Time to reach ground t= 2.44 s

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