A worker is being raised in a bucket lift at a constant speed of 3 fts . When the worker's arms are 10 ft off the ground, her coworker throws a measuring tape toward her. The measuring tape is thrown from a height of 6 ft with an initial vertical velocity of 15 ft s. Projectile motion formula: \( h=3 t+10 \) and \( \mathrm{h}=-16 \mathrm{t}^{2}+15 \mathrm{t}+6 \) and \( \mathrm{h}=-16 \mathrm{t}^{2}+6 \mathrm{t}+15 \) \( \mathrm{~h}=-16 \mathrm{t}^{2}+\mathrm{vt}+\mathrm{h}_{0} \) \( t= \) time, in seconds, since the measuring tape was thrown \( h= \) height, in feet, above the ground

To solve this problem, we need to find the time it takes for the measuring tape to reach the worker's arms. We can use the projectile motion formula to find the time. Given: - Initial height of the worker's arms: 10 ft - Initial height of the measuring tape: 6 ft - Initial vertical velocity of the measuring tape: 15 ft/s - Constant speed of the bucket lift: 3 ft/s We have two equations for the height of the measuring tape: 1. \( h = 3t + 10 \) 2. \( h = -16t^2 + 15t + 6 \) We need to find the time \( t \) when the measuring tape reaches the worker's arms, which is at a height of 10 ft. Let's solve the equations to find the time. Solve the equation by following steps: - step0: Solve for \(t\): \(3t+10=10\) - step1: Move the constant to the right side: \(3t=10-10\) - step2: Subtract the terms: \(3t=0\) - step3: Rewrite the expression: \(t=0\) Solving the equation \(3t + 10 = 10\) gives us \(t = 0\). This means that the measuring tape was thrown at time \(t = 0\). Now, let's solve the second equation to find the time when the measuring tape reaches the worker's arms. Solve the equation(The complex numbers system) by following steps: - step0: Solve using the quadratic formula in the complex numbers system: \(-16t^{2}+15t+6=10\) - step1: Move the expression to the left side: \(-16t^{2}+15t-4=0\) - step2: Multiply both sides: \(16t^{2}-15t+4=0\) - step3: Solve using the quadratic formula: \(t=\frac{15\pm \sqrt{\left(-15\right)^{2}-4\times 16\times 4}}{2\times 16}\) - step4: Simplify the expression: \(t=\frac{15\pm \sqrt{\left(-15\right)^{2}-4\times 16\times 4}}{32}\) - step5: Simplify the expression: \(t=\frac{15\pm \sqrt{-31}}{32}\) - step6: Simplify the expression: \(t=\frac{15\pm \sqrt{31}\times i}{32}\) - step7: Separate into possible cases: \(\begin{align}&t=\frac{15+\sqrt{31}\times i}{32}\\&t=\frac{15-\sqrt{31}\times i}{32}\end{align}\) - step8: Simplify the expression: \(\begin{align}&t=\frac{15}{32}+\frac{\sqrt{31}}{32}i\\&t=\frac{15-\sqrt{31}\times i}{32}\end{align}\) - step9: Simplify the expression: \(\begin{align}&t=\frac{15}{32}+\frac{\sqrt{31}}{32}i\\&t=\frac{15}{32}-\frac{\sqrt{31}}{32}i\end{align}\) - step10: Rewrite: \(t_{1}=\frac{15}{32}-\frac{\sqrt{31}}{32}i,t_{2}=\frac{15}{32}+\frac{\sqrt{31}}{32}i\) - step11: Remove the complex number(s): \(\textrm{No real solution}\) There is no real solution for the equation \(-16t^2 + 15t + 6 = 10\). This means that the measuring tape never reaches the worker's arms at a height of 10 ft. Therefore, the measuring tape does not reach the worker's arms at a height of 10 ft.