Hi I am new here so here is my question.

A gun fires a projectile at 2200 ft/s muzzle velocity at a target 2.0 km away, at the same altitude; neglecting wind, what angle above the horizon must you aim, to hit the target? acceleration due to gravity is 9.8 m/s^2.

Ty for any help.

Also where do I sign up? Sorry, but I am having trouble seeing where things are right now (posting this from a voice recognition engine).

3rd question. I am confused about this Y intercept stuff. Can someone please explain it to me?

(if you can't answer the last one, it's ok I really just need an answer for the first question). Ty so much if anyone can help ☺.

Guest Feb 24, 2015

#1**+10 **

For the horizontal component of motion the velocity is constant, so if the initial angle is θ, we have, using distance = velocity* time, and that 1ft/s = 0.3048m/s, that:

2000 = 670.56*cos(θ)*t ...(1) where t is the flight time.

For the vertical motion we have vertical distance = initial velocity*time + (1/2)*accn*time^{2}

Now the net vertical distance here is zero, so:

0 = 670.56*sin(θ)*t - (1/2)*9.8*t^{2 }...(2)

Rearrange equation (1) to get t = 2000/(670.56*cos(θ)) and substitute this into (2):

0 = 670.56*sin(θ)*2000/(670.56*cos(θ)) - (1/2)*9.8*(2000/670.56*cos(θ))^{2}

Multiply this through by cos(θ)^{2} and divide through by 2000 to get:

0 = cos(θ)*sin(θ) - (1/2)*9.8*2000/670.56^{2}

or 0 = (1/2)sin(2θ) - (1/2)*9.8*2000/670.56^{2}

Multiply through by 2 and add 9.8*2000/670.56^{2} to both sides:

sin(2θ) = 9.8*2000/670.56^{2}

so 2θ = sin^{-1}(9.8*2000/670.56^{2})

or θ = sin^{-1}(9.8*2000/670.56^{2})/2

$${\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{2\,000}}}{{{\mathtt{670.56}}}^{{\mathtt{2}}}}}\right)}}{{\mathtt{2}}}} = {\mathtt{1.249\: \!139\: \!800\: \!311\: \!5}}$$

θ ≈ 1.25°. A very small angle!

To sign up go to the Home page: http://web2.0calc.com/ and next to the Login button at the top right of the page you will be able to select *Register*.

.

Alan
Feb 24, 2015

#1**+10 **

Best Answer

For the horizontal component of motion the velocity is constant, so if the initial angle is θ, we have, using distance = velocity* time, and that 1ft/s = 0.3048m/s, that:

2000 = 670.56*cos(θ)*t ...(1) where t is the flight time.

For the vertical motion we have vertical distance = initial velocity*time + (1/2)*accn*time^{2}

Now the net vertical distance here is zero, so:

0 = 670.56*sin(θ)*t - (1/2)*9.8*t^{2 }...(2)

Rearrange equation (1) to get t = 2000/(670.56*cos(θ)) and substitute this into (2):

0 = 670.56*sin(θ)*2000/(670.56*cos(θ)) - (1/2)*9.8*(2000/670.56*cos(θ))^{2}

Multiply this through by cos(θ)^{2} and divide through by 2000 to get:

0 = cos(θ)*sin(θ) - (1/2)*9.8*2000/670.56^{2}

or 0 = (1/2)sin(2θ) - (1/2)*9.8*2000/670.56^{2}

Multiply through by 2 and add 9.8*2000/670.56^{2} to both sides:

sin(2θ) = 9.8*2000/670.56^{2}

so 2θ = sin^{-1}(9.8*2000/670.56^{2})

or θ = sin^{-1}(9.8*2000/670.56^{2})/2

$${\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{2\,000}}}{{{\mathtt{670.56}}}^{{\mathtt{2}}}}}\right)}}{{\mathtt{2}}}} = {\mathtt{1.249\: \!139\: \!800\: \!311\: \!5}}$$

θ ≈ 1.25°. A very small angle!

To sign up go to the Home page: http://web2.0calc.com/ and next to the Login button at the top right of the page you will be able to select *Register*.

.

Alan
Feb 24, 2015