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A rock is thrown vertically upward from ground level at time t = 0. At time t = 0.9 s it passes the top of a tall tower, and 1.5 s later it reaches its maximum height. What is the height of the tower? m

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Let the initial velocity be u

At maximum height velocity will be 0

Time taken to reach the maximum height=0.9+1.5=2.4 sec

v=u+at

0=u-(9.8)(2.4)

u=23.52 m/s

So Height of the tower =ut+1/2at^2

=(23.52)(0.9)-1/2(9.8)(0.9^2)

=17.199 m

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