Questions: A coin, thrown upward at time t=0 from an office in the Empire State Building, has height in feet above the ground t seconds later given by h(t)=-16 t^2+48 t+448=-16(t-7)(t+4) (a) From what height is the coin thrown? Include units in your answer. (b) At what time does the coin hit the ground? Include units in your answer.

Solution Steps

To determine the initial height from which the coin is thrown, evaluate the height function at the initial time. To find when the coin hits the ground, solve the height function for when the height is zero, considering only the positive time value since time cannot be negative.

To find the initial height from which the coin is thrown, we evaluate the height function \( h(t) \) at \( t = 0 \): \[ h(0) = -16(0)^2 + 48(0) + 448 = 448 \text{ feet} \]

To find the time when the coin hits the ground, we solve the equation \( h(t) = 0 \): \[ -16t^2 + 48t + 448 = 0 \] Solving this quadratic equation, we get: \[ t = \frac{-48 \pm \sqrt{48^2 - 4(-16)(448)}}{2(-16)} \] \[ t = \frac{-48 \pm \sqrt{2304 + 28672}}{-32} \] \[ t = \frac{-48 \pm \sqrt{30976}}{-32} \] \[ t = \frac{-48 \pm 176}{-32} \] This gives us two solutions: \[ t = \frac{-48 + 176}{-32} = -4 \quad \text{(not valid since time cannot be negative)} \] \[ t = \frac{-48 - 176}{-32} = 7 \] Thus, the coin hits the ground at \( t = 7 \) seconds.

Final Answer