Answer
(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 $17.0~m/s$
Work Step by Step
(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 $17.0~m/s$
