Question 193830
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Technically speaking, you cannot answer this question from the information given.  That is because you haven't actually given the initial velocity.  Yes, I know that you said the initial velocity is 55 ft/sec, but that is only part of the story.  55 ft/sec is a speed, i.e. a scalar quantity.  Velocity is a vector quantity and requires that you specify both the magnitude (speed) and the <u>direction</u>.  55 ft/sec North, 55 ft/sec horizontally, and 55 ft/sec West at a 45 degree angle to the horizontal, are all very different velocities.

Having said all that, I suspect that you really meant that the initial velocity is 55 ft/sec <u>up</u>.  So, let's solve the problem using that assumption.

The acceleration due to the force of gravity on an object near the earth's surface is

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ g = -32 \text{ \frac{ft}{sec^2}]

Note the minus sign because the accelleration vector is opposite in direction to our velocity vector.

Integrating with respect to time gives the instantaneous velocity at time <i><b>t</b></i>:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(t) = \int\,g\,dt = \int\,-32 \text{ \frac{ft}{sec^2}\,dt = -32t + C \text{ \frac{ft}{sec}}]

Where, for this problem, the constant of integration, <i><b>C</b></i>, is the initial velocity, and we will call this *[tex \LARGE v_o].  (If the velocity was in a direction other than straight up, then C would be the vertical component of the initial velocity)

Then integrating the velocity function with respect to time gives the instantaneous height at time <i><b>t</b></i>:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ H(t) = \int\,V(t) = \int\,-32t + v_o \text{ \frac{ft}{sec}} = -16t^2 + v_ot + C \text{ ft}]

Where the constant of integration is the initial height which we will call *[tex \LARGE h_o].  Hence the final function is:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ H(t) = -16t^2 + v_ot + h_o \text{ ft}]

And substituting the given initial values:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ H(t) = -16t^2 + 55t + 2 \text{ ft}]

Now the problem is to find *[tex \LARGE H(2)]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ H(2) = -16(2)^2 + 55(2) + 2 \text{ ft}]

The only thing left is a little arithmetic.

John
*[tex \LARGE e^{i\pi} + 1 = 0]
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