Question

A bullet fired from rifle attains a maximum height of 5m and crosses a range of 200m.find the angle of projection.

Answer 1

Answer:

$tan^{-1} (\frac{1}{10} )$

Explanation:

x = angle of projection

hmax = $\frac{(usinx)^{2} }{2g}$

$(usinx)^{2}$ = 100

usinx=10

range = ucosx * time of flight

time of flight = $\frac{2usinx}{g}$ = 2s

ucosx =  100

tan x= $\frac{1}{10}$

x = $tan^{-1} (\frac{1}{10} )$

Answer 2

Answer:

$\displaystyle \theta=\tan^{-1}\left(\frac{1}{10}\right)$

Explanation:

Given :

Height ( h ) = 5 m

Range ( R ) = 200 m

We have to find angle

We know ratio of Range and height

$\displaystyle \frac{R}{H}= \frac{4}{\tan\theta}$

Now put the value here

$\displaystyle \frac{200}{5}= \frac{4}{\tan\theta} \\\\\displaystyle \text{ \frac{\cancel {200}}{\cance {5}}= \frac{4}{\tan\theta} }\\\\\displaystyle \frac{40}{1}= \frac{4}{\tan\theta} \\\\\displaystyle \tan\theta= \frac{4}{40} \\\\\displaystyle \text{ \tan\theta= \frac{\cancel{4}}{\cancel{40}} }\\\\\displaystyle \tan\theta= \frac{1}{10} \\\\\displaystyle \theta=\tan^{-1}\left(\frac{1}{10}\right)$

Value is about to 0 as its decimal value is 0.099.

Thus we get answer.