Question

You throw a small rock straight up with an initial speed V0 from the edge of the roof of a building that is a distance H above the ground. The rock travels upward to a maximum height in time Tmax, misses the edge of the roof on its way down, and reaches the ground in time Ttotal after it was thrown. Neglect air resistance. If the total time the rock is in the air is three times the time it takes to reach its maximum height, so Ttotal = 3Tmax, then in terms of H, what must be the value of V0?

Answer 1

The value of
`\(V_0\)` must be:

`\[V_0 = (2)/(3)√(gH)\]`

Let's break down the problem step by step:

1. The rock is thrown straight up from the edge of the roof of a building, a distance H above the ground.

2. The rock travels upward to a maximum height and then falls back down.

3. The total time the rock is in the air is
`\(T_{\text{total}}\), and it's given that \(T_{\text{total}} = 3T_{\text{max}}\), where \(T_{\text{max}}\)`is the time it takes to reach the maximum height.

We want to find the value of V0, the initial speed of the rock, in terms of H.

Let's analyze the motion of the rock:

1. The time it takes to reach the maximum height is
`\(T_{\text{max}}\).`During this time, the rock is moving upward against gravity, and its velocity decreases until it reaches zero at the maximum height.

2. The time it takes to return from the maximum height to the ground is also
`\(T_{\text{max}}\)` because the motion is symmetric.

3. Therefore,
`\(T_{\text{total}} = 2T_{\text{max}}\).`

Now, we're given that
`\(T_{\text{total}} = 3T_{\text{max}}\),` so we can set up the equation:

`\[2T_{\text{max}} = 3T_{\text{max}}\]`

4. Now, we can solve for

`\[2T_{\text{max}} = 3T_{\text{max}}\]`

`$T_(\max )=(2)/(3) T_(\max )$`

5. Cancel out
`\(T_{\text{max}}\)` from both sides:

`\[1 = (2)/(3)\]`

6. Now, let's examine the kinematic equation for the motion of the rock:

`\[H = (1)/(2)gt^2\]`

Where g is the acceleration due to gravity and t is the time it takes to reach the maximum height, which is
`\(T_{\text{max}}\).`

7. Rearrange the equation to solve for V0:

`\[H = (1)/(2)g(T_{\text{max}})^2\]`

`\[T_{\text{max}} = \sqrt{(2H)/(g)}\]`

8. Now, plug in the expression for
`\(T_{\text{max}}\)` into the equation we found earlier:

`\[1 = (2)/(3)\]`

`\[(2)/(3) = \sqrt{(2H)/(g)}\]`

9. Square both sides to eliminate the square root:

`\[\left((2)/(3)\right)^2 = (2H)/(g)\]`

10. Solve for V0:

`\[(4)/(9) = (2H)/(g)\]`

`\[V_0 = \sqrt{(4)/(9)gH}\]`