A rock is thrown...?

Question

vertically upward from ground level at time t = 0. At time t = 1.6 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?

Answer 1

using
y(t)=v0*t-.5*g*t^2
and v(t)=v0-g*t

you can solve this problem.

At max height, v(t)=0, using g=10 m/s^2
0=v0-10*3.1
v0=31 m/s

with this now,
y(1.6) is the height of the tower,
y(1.6)=31*1.5-5*1.5^2
35.25 m

j

j