A player passes a basketball to another player who catches it at the same level from which it was thrown. The initial speed of the ball is 7.1m/s, and it travels a distance of 4.6m. What were (a) the initial direction of the ball and (b) the time of flight?

I can’t figure a way using only kinematic equations and soh cah toa, am I missing something? I tried using trigonometric identities but got stumped late into the algebra.

If $g$ denotes the gravity, $u$ the initial speed, $d$ the travelled horizontal distance, $xin left[0,fracpi2right]$ the angle to the ground of the initial velocity, and $t$ the flight time, then you have $u cos(x) t= d$ and $usin(x) t-frac12gt^2=0$. Solve for $x$ to get $sin(2x)=frac{gd}{u^2}$. There are two possible values for $x$, whence also two values for $t=frac{d}{ucos(x)}$.

Is the ball is thrown along a parabolic path at an angle $ alpha $ with speed $v$, the time is

$$ frac{distance }{v cos alpha} $$

So $alpha$ should also be known.