Question

A machine gun is mounted on the top of tower 100m high.At what angle should the gun be inclined to cover a max range of firing the ground below?Muzzle speed=150m/s,take g=10m/s^2

Answer 1

@siddhantsharan @RaphaelFilgueiras

Answer 2

@AravindG

Answer 3

range \[x=u _{x}t\cos (\theta)\]

Answer 4

time taken for the journey
\[u _{y}=u*\sin(\theta)\]
\[=150(\sin \theta)\]

Answer 5

Nop...@RaphaelFilgueiras ... U SEE..IT IS PROJECTED 100 M ABOVE...

Answer 6

answer is 43.78'

Answer 7

\[y=ut+\frac{ 1 }{ 2 }at^2\]
\[100=150(\sin \theta)t+(1/2)10t^2\]

Answer 8

use tan(theta)= ( 1+ 2gh/v^2)^(-1/2)

Answer 9

now...how did i get this formula for maximum range...
use equation of trajectory...
y = h+ x tan(theta) - gx^2/(2v^2) sec^2(theta)

Answer 10

for x= R(range), y=0

Answer 11

form a quadratic equation in tan(theta)

Answer 12

then use discriminant>=0 for real tan(theta)

Answer 13

then u'll get a relation
R<= v^2/g ( 1+ 2gh/v^2)^(1/2)

Answer 14

now try by ur own...

Answer 15

one more step...
put the value of Rmax in the quadratic equation of tan(theta)
then u will get the value of theta for which range will be maximum.

Answer 16

angular manipulation not looking nice
http://en.wikipedia.org/wiki/Range_of_a_projectile