
Projectile Motion
Here is a problem which i encountered, and have difficulties in terms of trying to solve.
Question: A tall building stands on level ground. The nozzle of a water sprinkler is positioned point P on the ground at a distance d from a wall of the building. Water sprays from nozzle with speed V and the nozzle can be pointed in any direction from P.
a) If V > sqrt (gd) prove that the water can reach the wall above ground level.
My attempt:
In this question, this is what i tried using standard vertical and horizontal motion, where t is time parameter, which i assumed was the approach as this is the current topic i am studying.
t = d/(Vcosa) where a is varying angle.
This is subbed into y = 1/2 gt^2  Vtsin a
Ultimately Ive obtained
gd tan^2 a  2V^2 tana + gd = y
Here by using quadratic formula, for y=0 to exist, V > sqrt (gd)
However, this solves for tan a, and i just cannot make the connection between V>sqrt gt and y is able to reach wall above ground level.
I assume this solution would depend upon inequality of
gd tan^2 a  2V^2 tana + gd > 0,
But this still solves for tan a. Is my method flawed, as I am thinking there may be another way without using tan a?

since water from the hose is projected from ground level, the max horizontal range is achieved when the launch angle = 45 degrees, so $\displaystyle \tan(a) = \tan(45) = 1
$
also, $\displaystyle y = v\sin(a) \cdot t  \frac{1}{2}gt^2$ (you had the signs reversed)
sub in $\displaystyle \frac{d}{v\cos(a)}$ for $\displaystyle t$ ...
$\displaystyle y = d\tan(a)  \frac{gd^2}{2v^2}\tan^2(a)  \frac{gd^2}{2v^2}$
since $\displaystyle tan(a) = tan(45) = 1$ ...
$\displaystyle y = d  \frac{gd^2}{v^2}$
for $\displaystyle v = \sqrt{gd}$ ...
$\displaystyle y = d  \frac{gd^2}{gd} = d  d = 0$
for $\displaystyle v > \sqrt{gd}$ , $\displaystyle d > \frac{gd^2}{v^2}$ which implies $\displaystyle d > 0$