Question

A rock is thrown vertically upward from ground level at time t = 0. At t = 1.6 s it passes the top of a tall tower, and 1.0 s later it reaches its maximum height. What is the height of the tower?

Answer 1

The height of the tower can be calculated by finding the maximum height reached by the rock and then subtracting the initial height of the rock.

Step-by-step explanation:

The height of the tower can be calculated by finding the maximum height reached by the rock and then subtracting the initial height of the rock. Since the rock reaches its maximum height 1.0 s after passing the top of the tower, we can use the formula:

Max height = y1 - y0 - (1/2) * g * t^2

where y1 is the maximum height, y0 is the initial height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it takes to reach the maximum height.

Since the rock is 8.10 m above its starting point at t = 1.00 s, we can plug in the values:

Max height = 8.10 m - 0 m - (1/2) * (9.8 m/s^2) * (1.0 s)^2

Simplifying this equation, we find that the maximum height reached by the rock is 3.15 m. Therefore, the height of the tower is 8.10 m + 3.15 m = 11.25 m.

Answer 2

The rock has height
`y` at time
`t` according to

`y=v_0t-\frac g2t^2`

where
`v_0` is the velocity with which it was thrown, and g = 9.8 m/s^2 is the acceleration due to gravity.

Complete the square to get

`y=\frac{{v_0}^2}{2g}-\frac g2\left(t-\frac{v_0}g\right)^2`

which indicates a maximum height of
`\frac{{v_0}^2}{2g}` occurs when
`t=\frac{v_0}g`. We're told this time is 2.6 s after the rock is thrown:

`2.6\,\mathrm s=(v_0)/(9.8(\rm m)/(\mathrm s^2))\implies v_0=25.48(\rm m)/(\rm s)`

So when t = 1.6 s, the rock reaches the tower's height of

`y=v_0(1.6\,\mathrm s)-\frac g2(1.6\,\mathrm s)^2\approx\boxed{28\,\mathrm m}`