An object is thrown straight up with a velocity in ft/s given by v(t)=-32t+67 where tis in seconds, from a height of 42 feet. a) What is the object's initial velocity? 67ft/s b) What is the object's maximum velocity? 67ft/s ft c) What is the object's maximum displacement? square square d) When does the maximum displacement occur? (67)/(32) e) When is the object's displacement 0? square f) What is the object's maximum height? square ft Note: You can earn partial credit on this problem.

c) To find the object's maximum displacement, we need to find the time when the velocity is 0. We can do this by setting the velocity equation equal to 0 and solving for t:<br /><br />$-32t + 67 = 0$<br /><br />$32t = 67$<br /><br />$t = \frac{67}{32}$<br /><br />Now, we can plug this value of t back into the displacement equation to find the maximum displacement:<br /><br />$x(t) = -16t^2 + 67t + 42$<br /><br />$x(\frac{67}{32}) = -16(\frac{67}{32})^2 + 67(\frac{67}{32}) + 42$<br /><br />$x(\frac{67}{32}) = -16(\frac{4489}{1024}) + 67(\frac{67}{32}) + 42$<br /><br />$x(\frac{67}{32}) = -71.40625 + 141.09375 + 42$<br /><br />$x(\frac{67}{32}) = 111.6875$<br /><br />Therefore, the object's maximum displacement is 111.6875 feet.<br /><br />d) The maximum displacement occurs at the time when the velocity is 0, which we found to be $\frac{67}{32}$ seconds.<br /><br />e) To find when the object's displacement is 0, we need to solve the displacement equation for t:<br /><br />$x(t) = -16t^2 + 67t + 42$<br /><br />$-16t^2 + 67t + 42 = 0$<br /><br />Using the quadratic formula, we get:<br /><br />$t = \frac{-67 \pm \sqrt{67^2 - 4(-16)(42)}}{2(-16)}$<br /><br />$t = \frac{-67 \pm \sqrt{4489 + 2688}}{-32}$<br /><br />$t = \frac{-67 \pm \sqrt{7177}}{-32}$<br /><br />$t = \frac{-67 \pm 84.7}{-32}$<br /><br />The two solutions are:<br /><br />$t_1 = \frac{-67 + 84.7}{-32} = -0.6$ (not physically meaningful since time cannot be negative)<br /><br />$t_2 = \frac{-67 - 84.7}{-32} = 4.6$<br /><br />Therefore, the object's displacement is 0 at $t = 4.6$ seconds.<br /><br />f) To find the object's maximum height, we need to find the time when the velocity is 0 and plug it into the displacement equation:<br /><br />$x(t) = -16t^2 + 67t + 42$<br /><br />$x(\frac{67}{32}) = -16(\frac{67}{32})^2 + 67(\frac{67}{32}) + 42$<br /><br />$x(\frac{67}{32}) = -71.40625 + 141.09375 + 42$<br /><br />$x(\frac{67}{32}) = 111.6875$<br /><br />Therefore, the object's maximum height is 111.6875 feet.