ectile \( \begin{array}{l}\text { 19. An object moving horizontally with speed } v \text { falls off } \\ \text { the edge of a vertical cliff and lands a distance } d \\ \text { from the base of the cliff. How far from the base of } \\ \text { the cliff would the object land if the height of the } \\ \text { cliff was doubled? } \\ \begin{array}{ll}\text { A) } d & \text { B) } \sqrt{2} d \\ \text { C) } 2 d & \text { D) } 2 \sqrt{2} d \\ \text { E) } 4 d & \end{array}\end{array}> \).

When an object falls off a cliff, its horizontal motion is independent of its vertical motion. The time it takes to hit the ground depends solely on the height of the cliff and is given by the equation \( t = \sqrt{\frac{2h}{g}} \), where \( h \) is the height and \( g \) is the acceleration due to gravity. If the height of the cliff is doubled, the time to fall becomes \( t = \sqrt{\frac{2(2h)}{g}} = \sqrt{2} \times \sqrt{\frac{2h}{g}} \). Thus, the object will spend more time in the air and travel further horizontally. If the horizontal speed \( v \) remains the same, and the fall takes \( \sqrt{2} \times t_1 \) (where \( t_1 \) is the original time), the new distance \( d' = v \times (\sqrt{2} \times t_1) = \sqrt{2} d \). Hence, the new distance from the base of the cliff is \( \sqrt{2} d \). So the answer is \( \text{B) } \sqrt{2} d \).