An object is thrown with a horizontal velocity of \( 15 \mathrm{~m} / \mathrm{s} \) off of a cliff that is 180 m above level ground. 33. If air resistance is negligible, the distance between the base of the cliff and the place where the object reaches the ground is most nearly \( 180 / 2 \) \( \begin{array}{ll}\text { A) } 50 \mathrm{~m} & \text { B) } 60 \mathrm{~m} \\ \text { C) } 150 \mathrm{~m} & \text { D) } 180 \mathrm{~m} \\ \text { E) } 90 \mathrm{~m} & \end{array} \)

To find the distance between the base of the cliff and where the object lands, we can use some physics equations. First, we need to determine how long it takes for the object to fall 180 m. The time \( t \) can be found using the equation for free fall: \[ d = \frac{1}{2} g t^2 \] where \( g \) is approximately \( 9.81 \mathrm{~m/s^2} \). Rearranging for \( t \): \[ 180 = \frac{1}{2} \cdot 9.81 \cdot t^2 \] \[ t^2 = \frac{180 \cdot 2}{9.81} \approx 36.71 \] \[ t \approx 6.06 \, \mathrm{s} \] Now, to find the horizontal distance \( d \): \[ d = v_x \cdot t = 15 \, \mathrm{m/s} \cdot 6.06 \, \mathrm{s} \approx 90.9 \, \mathrm{m} \] Rounding this gives us most nearly \( 90 \, \mathrm{m} \), which corresponds to option E. The answer is E) 90 m.