# Projectile Motion

• September 5th 2010, 05:41 AM
Lukybear
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?
• September 5th 2010, 06:57 AM
skeeter
since water from the hose is projected from ground level, the max horizontal range is achieved when the launch angle = 45 degrees, so $\tan(a) = \tan(45) = 1
$

also, $y = v\sin(a) \cdot t - \frac{1}{2}gt^2$ (you had the signs reversed)

sub in $\frac{d}{v\cos(a)}$ for $t$ ...

$y = d\tan(a) - \frac{gd^2}{2v^2}\tan^2(a) - \frac{gd^2}{2v^2}$

since $tan(a) = tan(45) = 1$ ...

$y = d - \frac{gd^2}{v^2}$

for $v = \sqrt{gd}$ ...

$y = d - \frac{gd^2}{gd} = d - d = 0$

for $v > \sqrt{gd}$ , $d > \frac{gd^2}{v^2}$ which implies $d > 0$