A stone 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.

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mariacsinning mariacsinning · Answer 1 · 2019-10-11T05:28:55+02:00

Answer:

The value of v0 is 65.11 ft/s

Explanation:

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

[tex]h_{max} =\frac{v_i^{2} }{2g}+h_i[/tex]

Where [tex]v_i[/tex] is the initial velocity of the stone, g is the gravity and [tex]h_i[/tex] 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 [tex]\frac{-g}{2}[/tex], so g is equal to [tex]32 ft/s^{2}[/tex]

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 [tex]h_{max}[/tex] of the first stone is:

[tex]h_{max} =\frac{36^{2} }{2*32}+46=66.25ft[/tex]

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

[tex]h_{max} =\frac{v_i^{2} }{2g}+h_0[/tex]

We can calculate [tex]v_0[/tex] as:

[tex]66.25 =\frac{v_0^{2} }{2*32}+0[/tex]

[tex]66.25=\frac{v_0^{2} }{64} \\ 66.25*64=v_0^{2} \\ \sqrt{4240} =v_0\\65.11=v_0[/tex]

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