A rock is thrown verticaly upward from ground level at time t = 0. At t = 1.65 s it passes the top of a tall tower, and 0.95 s later it reaches its maximum height. What is the height of the tower?

height = _______ m

First, find the initial velocity from the height of the tower:
Vf=vi - g time
where Vf is zero at the top.
solve for vi if t=1.65+.95

Then, knowing vi, solve for v at t=1.65

V=vi+ g t

To find the height of the tower, we need to break down the problem into two parts: finding the initial velocity of the rock and then using that to find the height of the tower.

1. Finding the initial velocity:
We know that at the top of the tower, the final velocity (Vf) of the rock is zero. We can use this information to find the initial velocity (vi) of the rock.

Using the equation Vf = vi - gt, where g is the acceleration due to gravity, we substitute in the values:
0 = vi - (9.8 m/s^2) * (1.65 + 0.95) s

Solving for vi:
vi = (9.8 m/s^2) * (1.65 + 0.95) s

2. Finding the height of the tower:
Now that we have the initial velocity (vi), we can find the time it takes for the rock to reach its maximum height. This will be the time it takes for the rock to travel upwards from the ground to its maximum height. Since the rock goes up and then comes back down, this time is half of the total time taken.

Using the equation V = vi + gt, we can solve for the time needed to reach the maximum height:
0 = vi - (9.8 m/s^2) * t_max_height

Solving for t_max_height:
t_max_height = vi / (9.8 m/s^2)

Now, using the time t_max_height, we can find the height of the tower using the equation for vertical motion:
height = vi * t_max_height - (1/2) * g * (t_max_height)^2

Substituting the values we have:
height = vi * (vi / (9.8 m/s^2)) - (1/2) * (9.8 m/s^2) * ((vi / (9.8 m/s^2))^2)

Simplifying the equation will give us the height of the tower in meters.