Arock is thrown straight up and passes by a window. the window is 1.7m tall, and the rock takes 0.19 seconds to pass from the bottom of the window to the top. how far above the top of the window will the rock rise?

The rock will rise 3.3 m above the top of the window.

Explanation:

The equations used to find the height and velocity of the rock at any given time are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the rock at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity

v = velocity of the rock at time t

If we place the frame of reference at the bottom of the window, we can say that at time t = 0.19 s the height of the rock is 1.7 m. That will allow us to find the initial velocity needed to find the time at which the rock is at its maximum height.

y = y0 + v0 · t + 1/2 · g · t²

1.7 m = 0 m + v0 · 0.19 s - 1/2 · 9.8 m/s² · (0.19 s)²

1.7 m + 1/2 · 9.8 m/s² · (0.19 s)²  = v0 · 0.19 s

(1.7 m + 1/2 · 9.8 m/s² · (0.19 s)²) / 0.19 s = v0

v0 = 9.9 m/s

With the initial velocity, we can find at which time the rock reaches its max- height. We know that at maximum height, the velocity of the rock is 0. Then, using the equation of velocity:

v = v0 + g · t

0 = 9.9 m/s - 9.8 m/s² · t

-9.9 m/s / -9.8 m/s² = t

t = 1.0 s

Now calculating the position at time t = 1.0 s, we will find the maximum heigth:

y = y0 + v0 · t + 1/2 · g · t²

y = 0 m + 9.9 m/s · 1.0 s - 1/2 · 9.8 m/s² · (1.0 s)²

y = 5.0 m  (this is the max-height meassured from the bottom of the window)

Then, the rock will rise (5.0 m - 1.7 m) 3.3 m above the top of the window.

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