The puck is sliding on the smooth floor with the speed of $V_{0} = 12 m / s$ , rises on a fixed frictionless wedge of height $h = 3.6 m$ and jumps from it. After jumping the puck lands on the other side of the wedge at a horizontal distance $S$ from the base of the wedge. The upper part of wedge is horizontal.

Question

TestHub · Accepted Answer

Concepts: Conservation of energy determines the puck's speed ( $v_{1}$ ) at the wedge top. Projectile motion equations then calculate the horizontal distance ( $S$ ).

Solution:

Energy conservation: $v_{1} = v_{0}^{2} - 2 g h = 144 - 72 = 62 m / s$ .

Projectile range: $S = v_{1} \frac{2 h}{g} = 62 \frac{7.2}{10} = 7.2 m$ .

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