Astone is thrown with an initial velocity of 36 ft/s from the edge of a bridge that is 46 ft above the ground. the height of this stone above the ground t seconds after it is thrown is f(t)= -16t^2 + 36t + 46. if a second stone is thrown from the ground, then its height above the ground after t seconds is given by g(t)= -16t^2 + v0t, where v0 is the initial velocity of the second stone. determine the value of v0 such that the two stones reach the same high point.

The value of v0 is 65.11 ft/s

Explanation:

The equation to find the highest point of every stone is:

Where is the initial velocity of the stone, g is the gravity and is the initial position of the stone.

From both equation: f(t)= -16t^2 + 36t + 46 and g(t)= -16t^2 + v0t, we can identify the following data:

1. The gravity: the number that is beside t^2 in both equations is equal to , so g is equal to

2. The initial velocity of the first stone: it is 36 ft/s, that is the number that is beside t in f(t)

3. The initial position of the first stone: it is 46 ft.

4. The initial position of the second stone: it is 0 ft

So, the of the first stone is:

If the second stone reach the same high point and the initial velocity for the second stone also satisfy the follow equation:

We can calculate as:

So, the value of of v0 such that the two stones reach the same high point is 65.11 ft/s

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