Maximum range angle for bullet fired from moving car at 20 m/s

[Suyash Singh]

As I indicated in post #23 there was a small ambiguity in Vriska's question.
Vriska means: if it looks to you in the cart as though it is going straight up above you then what will it look like to a bystander on the ground?

ok now i get what he meant by watching that video in #28

 

[kuruman]

R=u u sin(2θ)/g = u u 2 sinθ cosθ/g = (u u 2 voy/v vox/v)/g
= (u u 2 Vox Voy)/(g v v)
where vox and voy are x and y components of velocity wrt car.

No. For any projectile fired with initial speed $v_0$ relative to the ground, $$R=\frac{v_0^2\sin(2\theta)}{g}=\frac{2(v_0\sin \theta)(v_0 \cos\theta)}{g}=\frac{2v_{0y}v_{0x}}{g}$$
This equation gives the range if the projectile returns to the same vertical height from which it was fired. Now replace $v_{0x}$ and $v_{0y}$ with the expressions you found in post #20 and find $\theta$ such that $R$ is maximum.

 

[Suyash Singh]

R=2v0yv0x / g
= 2 (20cos(theta)+20)(20sin(theta))/g

now i use differentiation i suppose? to get 60 degrees

 

[kuruman]

now i use differentiation i suppose? to get 60 degrees

Yes. What expression did you get when you differentiated?

 

[Suyash Singh]

2cos(2theta)=-cos(theta)

2(cos^2θ-sin^2θ)=-cos(θ)

2(2cos^2θ-1)=-cos(theta)

then i solve it like a quadratic equation and then i differentiate it again to find maxima and minima

 

[kuruman]

That's it. Note that the quadratic gives two solutions. The maximum is at 60° as you already found. Also of interest is the minimum that occurs at -180° and gives zero range. It shows that if the pistol is fired horizontally and opposite to the direction of the car's motion, your friend will see the projectile just drop to the ground. That's the angle at which the projectile does not move horizontally, not straight up as you previously imagined. I hope all this is clear now.

 

[arvind12345]

.As the bullet has initial max range of 40 m so from there u can get value of u

Now R = [ucos (theta) +20] [2usin (theta)]/g [time of flight will not change]
= u x u sin2(theta)/g + 40 u sin(theta)/g
Now differentiate it with respect to theta to get the max range
I have not solved it. but i strongly feel that it will give max range for angle 45
For various different concepts refer Aavya classes Physics on you tube

 

[kuruman]

I have not solved it. but i strongly feel that it will give max range for angle 45

It is foolhardy to claim an answer by feelings alone without the algebra to back them up. In this case, the angle of inclination of the pistol for maximum range is not 45^o but 60^o.

 

[arvind12345]

It is foolhardy to claim an answer by feelings alone without the algebra to back them up. In this case, the angle of inclination of the pistol for maximum range is not 45^o but 60^o.

Thanks for reply. Ya u r right. Backing of algebra is needed.

 

[arvind12345]

It is foolhardy to claim an answer by feelings alone without the algebra to back them up. In this case, the angle of inclination of the pistol for maximum range is not 45^o but 60^o.

As per my understanding differentiation of below expression with respect to theta and keeping it equal to zero will give ans
= u x u sin2(θ)/g + 40 u sin(θ)/g
on differentiating and putting it to 0 we get finally
40 cos2θ +40 cosθ = 0
=> cosθ x cosθ -1 +cosθ=0
so its a quadratic eq which gives cosθ=1/2 and θ =60
=> so this is algebra to counter my feeling made in earlier post..

 

[kuruman]

Well done!