## University Physics with Modern Physics (14th Edition)

We can find the horizontal velocity. $v_x = v_0~cos(\theta) = (6.4~m/s)~cos(60^{\circ})$ $v_x = 3.2~m/s$ We can find the time $t$ for the coin to reach the dish. $t = \frac{x}{v_x} = \frac{2.1~m}{3.2~m/s} = 0.66~s$ We can find the initial vertical velocity. $v_{0y} = v_0~sin(\theta) = (6.4~m/s)~sin(60^{\circ})$ $v_{0y} = 5.54~m/s$ We can find the vertical component of velocity when the coin lands in the dish. $v_y = v_{0y} - gt$ $v_y = (5.54~m/s)- (9.80~m/s^2)(0.66~s)$ $v_y = -0.93~m/s$ We can find the height $h$ of the shelf. $h = \frac{v_y^2-v_{0y}^2}{2g} = \frac{(-0.93~m/s^2)^2-(5.54)^2}{(2)(-9.80~m/s^2)}$ $h = 1.5~m$ (a) The height of the shelf above the point where the coin leaves the hand is 1.5 meters. (b) The vertical component of velocity just before it lands in the dish is -0.93 m/s