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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

Best Answer 

 #1
avatar+26971 
+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*time2

Now the net vertical distance here is zero, so:

 

0 = 670.56*sin(θ)*t - (1/2)*9.8*t2   ...(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*2000/670.562

 

or 0 = (1/2)sin(2θ) - (1/2)*9.8*2000/670.562 

 

Multiply through by 2 and add 9.8*2000/670.562 to both sides:

 

sin(2θ) = 9.8*2000/670.562 

 

so 2θ = sin-1(9.8*2000/670.562)

 

or θ =  sin-1(9.8*2000/670.562)/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
avatar+26971 
+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*time2

Now the net vertical distance here is zero, so:

 

0 = 670.56*sin(θ)*t - (1/2)*9.8*t2   ...(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*2000/670.562

 

or 0 = (1/2)sin(2θ) - (1/2)*9.8*2000/670.562 

 

Multiply through by 2 and add 9.8*2000/670.562 to both sides:

 

sin(2θ) = 9.8*2000/670.562 

 

so 2θ = sin-1(9.8*2000/670.562)

 

or θ =  sin-1(9.8*2000/670.562)/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
 #2
avatar
+5

ty alan for the answer

Guest Feb 24, 2015
 #3
avatar+88871 
0

Thanks, Alan, for that nice answer.....

 

CPhill  Feb 24, 2015

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