Answer
$(a)\space 1.78\space s$
$(b)\space 20.8\space m/s$
Work Step by Step
(a) Let's apply equation 3.5b $S=ut +\frac{1}{2}at^{2}$ in the vertical direction to find the flight time of the golf ball.
$\uparrow S=ut +\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$-15.5\space m=0+\frac{1}{2}(-9.8\space m/s^{2})t^{2}$
$t^{2}=3.16\space s^{2}=>t=1.78\space s$
(b) Let's apply equation 3.3b $V=u+at$ in the vertical direction to find the vertical velocity of the golf ball.
$\uparrow V=u+at $ ; Let's plug known values into this equation.
$V=0+(-9.8\space m/s^{2})(1.78\space s)=-17.4\space m/s$
By using the Pythagorean theorem, we can get.
Speed of the ball $= \sqrt {V_{x}^{2}+V_{y}^{2}}=\sqrt {(11.4\space m/s)^{2}+(-17.4\space m/s)^{2}}=20.8\space m/s$
