Final Answer:
A block leaves a frictionless inclined surface horizontally after dropping off by a height h. Then the horizontal distance D where it will land on the floor, in terms of h, H, and g is a) D = √(2hH).
Step-by-step explanation:
When the block leaves the frictionless inclined surface, it acquires an initial velocity due to the gravitational potential energy it gained while falling a height h. This initial velocity can be decomposed into horizontal and vertical components. The horizontal component of velocity remains constant throughout the motion, as there is no horizontal force acting on the block.
The time of flight (the time it takes for the block to reach the floor) is determined by the vertical motion and can be expressed in terms of the initial height h and gravitational acceleration g. The formula for the time of flight is t = √(2h/g).
The horizontal distance D can be obtained by multiplying the horizontal component of velocity by the time of flight: D = v_horizontal * t. Substituting v_horizontal = √(2gh) and t = √(2h/g), we get D = √(2hH), where H is the initial height of the inclined surface.
Therefore, the correct answer is (a) D = √(2hH).