A shot fired from gun, elevation to hit a target 3000 feet away

[lightningkid99]

A shot fired from a gun with a muzzle velocity of 1200 ft per second is to hit a target 3000 feet away. Determine the minimum angle of elevation of the gun?

Ok so I know that r(t)= (vcos(theta) t)i +(vsin(theta) t +16t^2)j

I tried to set vcos(theta)t =3000, solve for t, and plug into the j component. However, I am getting confused with the math because i get t= 5/2cos(theta) and when I plug it in it gets pretty messed up. Can someone help?

 

[mfb]

It should not get too messy. Please show your calculations, otherwise it is impossible to tell what went wrong.
You either have a strange sign or angle convention with the j component.

 

[lightningkid99]

1200cos(theta) t=3000
t=3000/1200cos(theta) or 5/2cos(theta)

so I plug into the j component
1200sin(theta)(3000/1200cos(theta))- 16(5/2cos(theta))^2

That's where I'm stuck. I'm really not sure where to go from here

 

[mfb]

There are brackets missing for denominators, and the last expression should be equal to something. That allows to simplify it.

 

[James Salvadon]

A shot fired from a gun with a muzzle speed of 1200 feet per second is to hit a target 3000 feet away. Determine the minimum angle of elevation of the gun.

Work and answerYour mistake occurs because you wrote +16t^2 rather than -16t^2
r(t) = (Vocosθ)ti + ((Vosinθ)t-16t^2)j ( vector valued function)

Set the coefficient of the j component to 0 to get Vosinθt-16t^2 =0 when t=0 and t= (Vosint)/16

The range is given x= (Vocosθ)(Vosint)/16) = ((Vo)^2sin2θ))/32 , substituting using trig identity

x = (12002Sin2θ)/32 = 3000,

x= sin2θ = 1/15, θ = ½( sin^(-1)(1/15), θ ≈ 1.91 degrees

 

[PeroK]

A shot fired from a gun ...

This homework is over five years old. But, in any case, we don't give complete solutions to students here. See the homework guidelines:

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

 

[Frabjous]

homework is over five years old.

Doesn’t that make it a graduate reflection?

 

[Halc]

89.42 deg also works, technically not giving this now-grad his answer since the problem asked for a min angle, and my solution really really depends on a completely flat Earth in a vacuum (and a patient target). The solution otherwise depends significantly on the prevailing weather conditions.