If the projectiles follow the parabolic paths shown, who is hit first?

[fer Mnaj]

Homework Statement

[Moderator: Edited to remove political phrasing of a math problem.]

A ship fires simultaneously at two targets. If the projectiles follow the parabolic paths shown in the figure, which is hit first? Is there a way to solve it by using kinematic maths? I´m thinking if I solve the time for one projectile and then evaluate that exact time in the x(t) of the other could work, what you guys think? Is there enough information to solve?

Relevant Equations

x(t)=x0+v0xt
y(t)=y0+V0yt-1/2gt¨¨2

 

[haruspex]

I´m thinking if I solve the time for one projectile

Seems like a good start. Please post your attempt.

 

[haruspex]

I guess the speed is the same?

No need for any such assumption... and judging from the diagrams, not true.

 

[DaveC426913]

This is probably supposed to be solved mathematically, but a moment's look at the diagram will show intuitively which one will hit first.

Hint: ignore the horizontal component completely. In fact, make it zero (easily done: pretend you are behind the target craft in line with the shots, and thus can see no horizontal component at all).

Look at the vertical component only. Which one must "land" first?

 

[haruspex]

This is probably supposed to be solved mathematically, but a moment's look at the diagram will show intuitively which one will hit first.

Hint: ignore the horizontal component completely. In fact, make it zero. Look at the vertical component only. Which one must "land" first?

Hi Dave,

I replied privately to Peter because I was trying to avoid giving too much away. And I believe it is supposed to be solved the way you have.

 

[DaveC426913]

Oops.

That was silly of me to think this could be solved mathematically at all, given there are no numbers!
I guess I revealed more than intended by assuming the question was more complex than it actually is.

 

[haruspex]

I have been trying to solve it without assuming the same speed, but the best that I got was showing that ##u\geq v##, where ##u## is the speed of the projectile with lesser angle to the horizontal, if ##u\geq1##.
It made it more interesting but I am yet unable to solve it. What I have got was:
$$\frac{t_A}{t_B}=\underbrace{\frac{v}{u}}_{\leq 1}\left(\underbrace{\frac{\sin(\varphi+\theta)}{\sin(\varphi)}}_{\geq 1}\right)
=\underbrace{\frac{x}{x+\Delta x}}_{\leq 1}\left(\underbrace{\frac{u\cos(\varphi)}{v\cos(\varphi+\theta)}}_{\geq 1}\right)$$
I do not know how to proceed from this, though.

What aspect of the launch variables determines the time in the air?

 

[neilparker62]

$$x_a=vcos(θ_a).t_a$$
$$x_b=vcos(θ_b).t_b$$
$$\frac{t_a}{t_b}=\frac{x_acos(θ_b)}{x_bcos(θ_a)}$$

Assumes the initial velocities have the same magnitude. So if (for example) the respective angles were 60 and 30, the range ratio would be ##\sqrt3## for equal times.

 

[haruspex]

The vertical one,

Right. What in the diagram directly gives you information about how the two vertical speeds compare?

 

[haruspex]

Well, I have ##v_{Ay}=v\sin(\varphi+\theta)## and ##v_{By}=u\sin(\varphi)##. But, although ##\sin(\varphi+\theta)\geq\sin(\varphi+\theta)##, we also have ##u\geq v##.

You don't need any algebra. From the diagram you should immediately be able to see how the vertical components of the launch velocities compare.

 

[PeroK]

Well, I have ##v_{Ay}=v\sin(\varphi+\theta)## and ##v_{By}=u\sin(\varphi)##. But, although ##\sin(\varphi+\theta)\geq\sin(\varphi+\theta)##, we also have ##u\geq v##.

Whose homework is this anyway?

 

[archaic]

You don't need any algebra. From the diagram you should immediately be able to see how the vertical components of the launch velocities compare.

Yeah, I wanted to generalize.

 

[neilparker62]

$$x_a=vcos(θ_a).t_a$$
$$x_b=vcos(θ_b).t_b$$
$$\frac{t_a}{t_b}=\frac{x_acos(θ_b)}{x_bcos(θ_a)}$$

Assumes the initial velocities have the same magnitude. So if (for example) the respective angles were 60 and 30, the range ratio would be ##\sqrt3## for equal times.

Oops - just realized the time ratio CANNOT be unity. As others have mentioned we should be looking only at the vertical component of velocity and since in the vertical direction , there is no relative acceleration, the projectiles must continually separate: ## y_a-y_b = [vsin(θ_a)-vsin(θ_b)]t ##.

 

[kuruman]

To echo @haruspex's post #10. Just look at the drawing. The projectile that spends less time in the air hits the target first. Focus on the vertical motion and think maximum height.

 

[neilparker62]

To echo @haruspex's post #10. Just look at the drawing. The projectile that spends less time in the air hits the target first. Focus on the vertical motion and think maximum height.

Yes - if you look at the horizontal axis - apart from anything else - it is very misleading! Following is a screen shot where I (approximately) measure the projectile angles. Based on these angles an interesting exercise is to determine expressions for the respective distances to target A and target B in terms of a presumed common projectile velocity v and accn of gravity g.

 

[kuruman]

Yes - if you look at the horizontal axis - apart from anything else - it is very misleading! Following is a screen shot where I (approximately) measure the projectile angles. Based on these angles an interesting exercise is to determine expressions for the respective distances to target A and target B in terms of a presumed common projectile velocity v and accn of gravity g.

View attachment 257163

The drawing is misleading only if you assume that the muzzle velocity is the same for the two projectiles. That's because the initial angles appear to be close to equidistant from the 45^o angle which means that the ranges must be more or less equal. If we accept that the drawing is to scale, we have to relax the requirement of equal muzzle velocities. @haruspex hinted at that in #3.

 

[neilparker62]

The drawing is misleading only if you assume that the muzzle velocity is the same for the two projectiles. That's because the initial angles appear to be close to equidistant from the 45^o angle which means that the ranges must be more or less equal. If we accept that the drawing is to scale, we have to relax the requirement of equal muzzle velocities. @haruspex hinted at that in #3.

True enough - Haruspex had it right in all respects. For interest here's a Wolfram Alpha plot which more or less shows the situation in the diagram. Have chosen angles of 60 and 30, velocities 100 and 150 m/s and g = -10m/s^2. The angles look somewhat distorted because I can't get WA to do 1:1 axes scaling. As can be seen, the 60 degree projectile has some way to go at the time instant (15 s) when the 30 degree one hits its target.