Answer
$(a)\space 1380\space m/s$
$(b)\space 66^{\circ}$ below the horizontal
Work Step by Step
Please see the attached image first.
Let's apply equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in the vertical direction to find the flight time of the projectile.
$\uparrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$-732\space m=97.5\space m/s\times sin50^{\circ}t+\frac{1}{2}(-9.8\space m/s^{2})t^{2}$
$4.9t^{2}-74.7t-732=0$
This is a quadratic equation & we can use the quadratic formula to find the solution for t.
$t=\frac{-(-74.7)\space \pm \sqrt {(-74.7)^{2}-4(4.9)(-732)}}{2(4.9)}s=\frac{74.7\space \pm 141.16}{9.8}s$
$t=-6.78\space s\space or\space t=22\space s$
Since time is a positive value we can neglect the negative solution for the t. So we can get,
$t=22\space s$
Let's apply equation 3.5a $S=ut$ in the horizontal direction to find the horizontal displacement of the projectile.
$\rightarrow S=ut$ ; Let's plug known values into this equation.
$x=97.5\space m/s\times cos50^{\circ}\times 22\space s=1380\space m$
Let's apply equation 3.3b $V=u+at$ in the vertical direction to find the vertical velocity after 22 s.
$\uparrow V=u+at$ ; Let's plug known values into this equation.
$V_{y}=97.5sin50^{\circ}\space m/s+(-9.8)22\space m/s=-141\space m/s$
By using trigonometry we can get,
$tan\theta=\frac{141\space m/s}{62.7\space m/s}=>\theta=tan^{-1}(2.25)=66^{\circ}$ Below the horizontal.
