Question

A hunter in a valley is trying to shoot a deer on the hill. The distance of the deer along his line of sight is 10 multiplied by square root of 181 and the height of the hill is 90m. His gun has a muzzle velocity of 100m/s. How many meters above the deer should he aim his rifle in order to hit him

Answer 1

Final Answer:  1810m  or 10m 

Steps:
1)  By Pythagoras Theorem,
Base ,x = $\sqrt{ (10 \sqrt{181} )^{2} - (90)^{2} } = \sqrt{18100-8100} = 100 m$

Target (deer ) is at :
 y = 90m
 x =  100 m

2) Equation of Trajectory /Equation of Path : 

 $\:\:\:\:y=xtan(\theta)- \frac{gx^{2}}{2u^{2}cos^{2}(\theta)} \\ \\ =\ \textgreater \ 90 = 100tan(\theta)- \frac{10*100^{2}*(1+tan^{2}(\theta))}{2*100^{2}} \\ \\ =\ \textgreater \ 18=20tan(\theta) -(1+ tan^{2}(\theta)) \\ \\ =\ \textgreater \ tan^{2}(\theta)-20tan(\theta)+19=0\\ \\ =\ \textgreater \ (tan(\theta)-1)(tan(\theta)-19)=0\\ \\ =\ \textgreater \ tan(\theta) = 1 \: or\: tan(\theta)=19$

3) When  
$tan(\theta) = 1 \\ \\ =\ \textgreater \ \frac{h+90}{100} = 1 \\ \\ =\ \textgreater \ h =10m$ 

$tan(\theta) = 19\\ \\ =\ \textgreater \ \frac{h+90}{100} = 19 \\ \\ =\ \textgreater \ h =1810m$

4) Hence ,there are two possibilites : 
To hit the deer,hunter should aim his rifle  1810m or 10m above the deer to dircetly hit the deer. 

Answer 2

This is the answer..

Hope it helps..