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

Answer:

0.122 radians

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,

$s=u_{x}\times t$

Here, $u_{x}$ is horizontal velocity and t is time.

Substitute all the values in above equation, we get

$s=ucos\theta\times t\\200=90cos\theta\times t\\t=\dfrac{200}{90cos\theta}\\t=\dfrac{20}{9cos\theta}$

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

$s_{y}=u_{y}t+\frac{1}{2}at^{2}$

Substitute all the variables.

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

Therefore the required angle of elevation is 0.122 radians.