Projectile motion of Golf ball- Can projection height be calculated just from initial velocity and travel time? 
A Golf ball projected with a horizontal velocity of 30 meter per
  second and takes 4 second to reach the ground. Calculate the height
  the golf ball was projected from. Calculate the magnitude of golf
  ball's vertical velocity component just before hitting ground.

Is this problem solvable? Can I assume the projection angle was 45 degree?
I attempted as following:
Horizontal acceleration is zero so,
$$
v_x(t) = v_x(0),
$$
therefore the horizontal displacement is,
$$
\Delta x = v_x(0) t = 30 \cdot 4 = 120\ m.
$$
Vertical displacement is,
$$
\Delta y = y(t) - y(0) = v(0) \sin\theta - \frac{gt^2}{2} = 30 \sin\theta - 0.5 \cdot 9.8 \cdot 4^2 = 30 \sin\theta - 78.4. \tag{1}
$$
I suspect I am wrong in assuming $v(0) = 30\ m/s$ because the question only states the horizontal velocity.
If actually $v(0)$ is $30\ m/s$ then I can take it further as,
$$
v(0) = v_x(0) \cos\theta
$$
$$
30 = 30 \cos\theta
$$
$$
\cos\theta = 1
$$
$$
\theta = 0^\circ
$$
Then replacing $\theta$ in (1) gives $\Delta y = 78.4\ m$ which is definitely wrong.
 A: The ball was in flight for four seconds: we can safely say that the ball reached maximum height at $t = 2$. (The gravitational pull is constant and there are no other forces acting, so the flight path is symmetrical). The ball was stationary at $t=2$ so its speed is $=0$
So now use the formula $v= u + at$, where $a$= acceleration, $t$= time, $u$= initial speed, $v$= final speed:
$v= u + at$ 
$0 = u - (g)(2)$ 
therefore $u = 2g$
And so we have the initial upwards speed component.
We could calculate the velocity at which it lands, but it's easier to argue that  when it lands it will also land at the same speed due to conservation of energy.
To work out the maximum height you could use the formula:
$h = ut + (1/2)(a)(t^2)$ when $t=2$, where the variables represent the same quantities.
The maximum height gained is in fact $19.6m$ when you plug in $u=2g$, $a=9.8$ .
Note that the $30m/s$ is irrelevent since it is the HORIZONTAL speed which has no effect on the vertical speed.
A: $\Delta y=78.4m$ is the correct answer.
The golf ball was fired horizontally from a height of $78.4m$.
With $\Delta y= \frac{1}{2}gt^2$ and $t=4s$ we get $\Delta y=78.4m$.
The horizontal velocity component of $30m/s$ was a distraction, not needed for calculating the solution because the vertical component and horizontal component of velocity are completely independent of each other. 
The $30m/s$ would allow you to calculate how far the golf ball flew before hitting the floor but that wasn't part of the problem.
As regards the golf ball's vertical velocity component just before hitting the ground, it's given by $v_y=gt$, thus $v_y=39.2m/s$.

