## College Physics (4th Edition)

(a) The maximum height of the ball above the ground is $5.9~m$ (b) At the highest point, the speed of the ball is $17.0~m/s$
(a) We can find the vertical displacement above $1.0~m$ when the ball is at maximum height: $v_{yf}^2 = v_{0y}^2+2a\Delta y$ $\Delta y = \frac{v_{yf}^2 - v_{0y}^2}{2a}$ $\Delta y = \frac{0 - (19.6~m/s~sin~30.0^{\circ})^2}{(2)(-9.80~m/s^2)}$ $\Delta y = 4.9~m$ The maximum height of the ball above the ground is $5.9~m$ (b) At the highest point, the vertical component of velocity is zero. Therefore, the speed is equal to the magnitude of the horizontal component of velocity: $speed = v_x = (19.6~m/s)~cos~30.0^{\circ} = 17.0~m/s$ At the highest point, the speed of the ball is $17.0~m/s$