Question

physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor.

Answer 1

Explanation:

$t = 0.35 \\ t = \sqrt{ \frac{2h}{g} } \\ \frac{t \sqrt{g} }{ \sqrt{2} } = \sqrt{h} \\ h = \frac{g {t}^{2} }{2} \\ h = \frac{9.8 \times 0.1225}{2} \\ h = 0.60025m \\ r = ux \times t \\ ux = 1.1 \\ r = 0.35 \times 1.1 = 0.385m \\ let \: v \: be \: the \: velocity \: of \: the \: projectile \: \\ just \: before \: tuching \: the \: ground \\ then \: \: vx = ui = 1.1 \\ vy = uyi - gtj \\ now \\ vy = uy + gt \\ uy = 0 \\ vy = ( gt) \\ then \: the \: velociy \: of \: the \: projectile \\ = v = \sqrt{ {vx}^{2} + {vy}^{2} } \\ v = 10.902 \frac{m}{s} \\ dierection \: \tan( \alpha ) = \frac{vy}{vx} \\ \\ \tan( \alpha ) = 3.1181 \\ \alpha = { \tan }^{ - 1} (3.11811) \\ \alpha = 72.21$