Question

(1 point) A gun has a muzzle speed of 90 meters per second. What angle of elevation θ, where 0≤θ≤π4, should be used to hit an object 200 meters away? Neglect air resistance and use g=9.8m/sec2 as the acceleration of gravit

Answer 1

0.122 radians

Step-by-step explanation:

Given that the muzzle speed is,
`u=90m/s`

And the distance of the object is,
`s=200 m`

And acceleration due to gravity is
`g=9.8 m/s^(2)`

Therefore negative acceleration due to gravity for upward motion.

Now,

The x component of velocity will be,
`u_(x)=ucos\theta`

The y component of velocity will be,
`u_(y)=usin\theta`

And the horizontal displacement is given. Now horizontal displacement is,

`s=u_(x)* t`

Here,
`u_(x)` is horizontal velocity and t is time.

Substitute all the values in above equation, we get

`s=ucos\theta* t\\200=90cos\theta* t\\t=(200)/(90cos\theta)\\t=(20)/(9cos\theta)`

Now, in y direction we can calculate displacement and in y direction displacement is zero.

`s_(y)=u_(y)t+(1)/(2)at^(2)`

Substitute all the variables.

`0=usin\theta t+(1)/(2)at^(2)\\0=90sin\theta ((20)/(9cos\theta))+(1)/(2)(-9.8 m/s^(2))* ((20)/(9cos\theta))^(2) \\200tan\theta=4.9* (400)/(81)* (1)/(cos^(2)\theta)\\200sin\theta=(24.1975)/(cos\theta)\\ 2sin\theta cos\theta=0.241975\\sin2\theta=0.241975\\2\theta=sin^(-1)(0.241975)\\ \theta=0.122 radians`

Therefore the required angle of elevation is 0.122 radians.