A ball is launched horizontally at 20 m/s off of a cliff which is meters tall What is the vertical component of the ball's velocity 1.5 seconds later ? (negative answer)

What property is a characteristic of a substance that can be observed and does not change the identity of the substance

Initially, a particle is moving at 5.33 m/s at an angle of 37.9° above the horizontal. two seconds later, its velocity is 6.11 m/s at an angle of 54.2° below the horizontal. what was the particle's average acceleration during these 2.00 seconds in the x-direction (enter first) and the y-direction?

Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).