Question

A rock is thrown vertically upward from ground level at time t = 0. At time t = 1 s it passes the top of a tall tower, and 1 s later it reaches its maximum height. What is the height of the tower?
__ m

Answer 1

d = vf*t - 0.5*a*t^2
= (0)(>1.0) - (0.5)(-9.8)(1.1^2) (since after 1 sec it reaches max height so here t = 1.1 say)
= 59.29m

using max. height find inital speed of rock when thrown:

Vf^2 = Vi^2 + 2ad
0^2 = Vi^2 + (2)(-9.8)(59.29)
Vi = 34.089 m/s

now you can find height of tower:

d = vi*t + 0.5*a*t^2
= (34.089)(1) + (0.5)(-9.8)(1^2)
= 38.989 m