Question

A rock is thrown vertically upward from ground level at time t = 0. At time t = 1.4 s it passes the top of a tall tower, and 1.2 s later it reaches its maximum height. What is the height (in m) of the tower?

Answer 1

Total time for rock to fall to ground = 1.4+1.2+1.2+1.4 = 5.2

?y = 0 because it went up and back down to the same height

g= -9.8 m/s^2    acceleration due to gravity

?y = vt + 1/2gt^2

0 = v(5.2) + 0.5(-9.8)(5.2)^2

5.2v = 132.496

v = 25.48 m/s

Height of tower= vt + 0.5gt^2 = 25.48(1.4) + 0.5(-9.8)(1.4)^2 = 26.068 m

Good Luck on your studies

Answer 2

First consider the rock going from ground level to the max. height to get the intial velocity of the rock

0 = v0 -9.8*2.7 --> v0 = 26.46 m/s

the motion from theground to the top of the tower only is

y = v0t +(1/2)at^2

= 26.46*1.2 - 0.5*9.8*1.2^2

= 24.69 m