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Search the web... tps://leam.vccs.edu/bbcswebdav/pid-114940965-dt-content-rid-243548110 2 ss NorthernVir. ) Virginia Community Colleges..-PHY2010Lecture Notes-NV2. tjsp-course id... Amazon.com-Online Sh.... Priceline.com .20 1.060Visu 18/Physics%20201 %28Chapter learn vccs edu Example 2-10: Falling from tower: Suppose a ball is dropped and thrown downward from a tower of 70 m height with an initial velocity of 3 m/s What is the position of the ball after 1 and 2 sec? and what would be its speed after 1 and 2 second? iD

Explanation / Answer

Here,

initial velocity , v = 3 m/s

at 1 s

position of the ball = u *t + 0.50 * at^2

position of the ball = 3 * 1 + 0.50 * 10 * 1^2

position of the ball = 8 m

ball is 8 m below the top of tower

at t = 2 s

position of the ball = u *t + 0.50 * at^2

position of the ball = 3 * 2 + 0.50 * 10 * 2^2

position of the ball = 26 m

position of the ball is 26 m below the tower

===========================================

at t = 1 s

speed = 1 * 10 + 3 = 13 m/s

the speed is 13 m/s at 1 s

at t= 2 s

speed = 2 * 10 + 3 = 23 m/s

the speed is 23 m/s

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