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Constant Acceleration Question

  • Thread starter phyphyphy
  • Start date
3
0
A rock is thrown vertically upward from ground level at time t = 0. At t = 1.6 s it passes the top of a tall tower, and 2.0 s later it reaches its maximum height. What is the height of the tower?

My thought process:

First find initial velocity:

v=vo+at
0=vo+(-9.8)(2)
vo=19.6 m/s

Then you can find the height of the tower, where x-xo is the displacement/height of the tower:
x-xo=vo*t+.5*a*t^2
x-xo=19.6*1.6+.5*-9.8*1.6^2
x-xo=18.816 m

However, WileyPlus says that I am wrong. What am I doing wrong?
 

Answers and Replies

Nathanael
Homework Helper
1,650
238
0=vo+(-9.8)(2)
vo=19.6 m/s
You did almost everything correctly. But is it true that it's maximum height is reached after 2 seconds?
(That's what your math says.)
 
3
0
OH, so I should have the maximum height as 3.6 seconds, correct?
 
Nathanael
Homework Helper
1,650
238
OH, so I should have the maximum height as 3.6 seconds, correct?
Correct.
 

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