3. (a) [3]moving object has elevation tion as a function of time given by h(t)=a+bt-ct^2 __ where a, b and care constants. What is the dimension of each constant? [a]= [b]= [e]= (b) [5] In the equation above, the values of the constants in FPS units are a=15,b=30 and c=18 Determine the values of hand tin FPS units, at the instant when the object reaches its maximum elevation.

**(a) Dimensional Analysis**<br /><br />* **[a]:** Since `a` is added to terms that represent elevation (a measure of distance), `a` must also have dimensions of length. Therefore, `[a] = L` (where L represents length).<br /><br />* **[b]:** The term `bt` must also have dimensions of length. Since `t` represents time, `b` must have dimensions of length divided by time. Therefore, `[b] = L/T` (where T represents time). This represents velocity.<br /><br />* **[c]:** The term `ct²` must also have dimensions of length. Since `t²` represents time squared, `c` must have dimensions of length divided by time squared. Therefore, `[c] = L/T²` (where T represents time). This represents acceleration.<br /><br /><br />**(b) Finding Maximum Elevation**<br /><br />The equation for elevation is given by:<br /><br />```<br />h(t) = a + bt - ct²<br />```<br /><br />In FPS units, we have a = 15 ft, b = 30 ft/s, and c = 18 ft/s².<br /><br />To find the time at which the object reaches its maximum elevation, we can take the derivative of h(t) with respect to time and set it equal to zero. This is because the slope of the tangent line to the curve of h(t) is zero at the maximum point.<br /><br />```<br />dh(t)/dt = b - 2ct<br />```<br /><br />Setting the derivative equal to zero:<br /><br />```<br />0 = b - 2ct<br />```<br /><br />Solving for t:<br /><br />```<br />t = b / (2c)<br />```<br /><br />Substituting the given values:<br /><br />```<br />t = (30 ft/s) / (2 * 18 ft/s²) = 30/36 s = 5/6 s<br />```<br /><br />Now, we can substitute this value of t back into the original equation for h(t) to find the maximum elevation:<br /><br />```<br />h(5/6) = 15 ft + (30 ft/s)(5/6 s) - (18 ft/s²)(5/6 s)²<br /> = 15 ft + 25 ft - (18 ft/s²)(25/36 s²)<br /> = 15 ft + 25 ft - 12.5 ft<br /> = 27.5 ft<br />```<br /><br />Therefore, the maximum elevation is 27.5 ft, and it occurs at t = 5/6 seconds.<br /><br /><br />**Final Answer:**<br /><br />(a) `[a] = L` , `[b] = L/T` , `[c] = L/T²`<br /><br />(b) Maximum elevation: h = 27.5 ft, Time: t = 5/6 s<br />