Projectile motion with (constant) wind velocity

[brotherbobby]

Homework Statement

A cannon ball fired like a projectile has a range $R$ on a still day. On a windy day, the wind blows horizontally at a constant speed. The ball fired with the same speed at the same angle relative to the cannon is found to have a range of $R+2H$ on the windy day, where $H$ is the maximum vertical height attained by the ball. How is the wind velocity related to the initial velocity of the ball relative to the cannon?

Relevant Equations

(1) Range of a projectile : $R = v_{0x} T$ where $v_{0x}$ is the initial velocity in the $x$ direction and $T$ is the total time of flight. (This range increases due to wind, implying that $v_{0x}' = v_{0x}+v_w$ where $v_w$ is the wind velocity. Crucially, the time(s) of flight remain the same).

(2) The time of flight $T = \dfrac{2v_{0y}}{g}$, which remains unaltered because the initial $\underline{\text{vertical}}$ component of the projectile velocity, $v_{0y}$ remains the same.

(3) The maximum height of the projectile $H = \frac{v^2_{0y}}{2g}$, which remains the same for the same reason as (2) above.

Let me start be making a small sketch of the problem, shown to the right.

If the range of the projectile on a still day $R = v_{0x}T$, then on the windy day the range becomes $R+2H = v'_{0x}T = (v_{0x}+v_w) T$.
Since the maximum height attained by the projectile $H = \dfrac{v_{0y}^2}{2g}$, the above equation simplifies to : $\cancel{v_{0x}T} + \dfrac{v_{0y}^2}{g} = (\cancel{v_{0x}}+v_w) T$. But since the time of flight $T = \dfrac{2v_{0y}}{g}$, we have $\dfrac{v_{0y}^\cancel{2}}{\cancel{g}} = \dfrac{2\cancel{v_{0y}}v_w}{\cancel{g}}\Rightarrow \boxed{v_w = \dfrac{v_{0y}}{2}}$.

A hint or help would be welcome if I am mistaken somewhere.

 

[PeroK]

Why do you think your answer might be wrong?

 

[brotherbobby]

Why do you think your answer might be wrong?

Yes good question. I was wondering how to relate the initial velocities in the two situations ($v_0$ and $v'_0$) and the angles in the two situations ($\theta_0$ to $\theta '_0$).

I have done neither. In retrospect, my final answer is not put correctly, because nowhere in the statement of the problem is $v_0$ or $v_{0y}$ mentioned. I can fix that, for now, without worrying about the earlier problem still undone.

We have as our answer $v_w = \frac{v_0y}{2}$. But $H = \frac{v^2_{0y}}{2g}\Rightarrow v_{0y} = \sqrt{2gH}$. Hence, $\boxed{v_w = \sqrt{\frac{gH}{2}}}$,which is in terms of the height H mentioned in the problem statement.

Question now is, how do the initial velocities and initial angles relate to one another? I will get back to you.

Please let me be aware of any comments or mistakes and thanks for your time.

 

[PeroK]

Yes good question. I was wondering how to relate the initial velocities in the two situations ($v_0$ and $v'_0$) and the angles in the two situations ($\theta_0$ to $\theta '_0$).

I have done neither. In retrospect, my final answer is not put correctly, because nowhere in the statement of the problem is $v_0$ or $v_{0y}$ mentioned. I can fix that, for now, without worrying about the earlier problem still undone.

We have as our answer $v_w = \frac{v_0y}{2}$. But $H = \frac{v^2_{0y}}{2g}\Rightarrow v_{0y} = \sqrt{2gH}$. Hence, $\boxed{v_w = \sqrt{\frac{gH}{2}}}$,which is in terms of the height H mentioned in the problem statement.

Question now is, how do the initial velocities and initial angles relate to one another? I will get back to you.

Please let me be aware of any comments or mistakes and thanks for your time.

What you've done looks right to me. This book answer is the same, but expressed in terms of $H$, rather than $v_{0y}$.

 

[brotherbobby]

@PeroK . I have found relations of $v_0$ to $v'_0$ and $\theta_0$ to $\theta'_0$. They are messy but I hope I am right.

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(1) The initial velocities :
Since the initial velocities are unaltered, we have $v_{0y} = v'_{0y}\Rightarrow v_0 \sin \theta_0 = v'_0 \sin\theta'_0$ . Let me save this as the first equation.
$$v_0 \sin \theta_0 = v'_0 \sin\theta'_0$$

Again, when there is wind, the initial velocity in the horizontal direction $v'_{0x} =\underbrace{v_0 \cos\theta_0}_{v_{0x}}+v_w = v'_0 \cos\theta'_0$. Saving this equation
$$v'_0 \cos\theta'_0=\underbrace{v_0 \cos\theta_0}_{v_{0x}}+v_w$$​

Squaring the two equations and adding, we obtain $v'^2_0= v^2_0 + v^2_0 \sin \theta_0 \cos \theta_0+v^2_0 \frac{\sin^2 \theta_0}{4}\Rightarrow \boxed{v'_0 = v_0\sqrt{1+\sin \theta_0 \cos \theta_0+\frac{\sin^2 \theta_0}{4}}}$.

(2) The initial angles :

Using the first saved equation $v_0 \sin \theta_0 = v'_0 \sin\theta'_0$ and the expression for the initial velocities above, we obtain : $$\boxed{\theta'_0 = \sin^{-1} \dfrac{\sin\theta_0}{\sqrt{1+\sin\theta_0 \cos\theta_0+\frac{\sin^2 \theta_0}{4}}}}$$

 

[PeroK]

I assumed the cannon fires at at fixed but unknown angle? The same angle both days?

 

[haruspex]

While this approach may be that intended, it is quite divorced from reality.
The direction of drag is opposite to that of the ball's air velocity and, at these speeds, varies as the square of that. So when there is a following wind the drag is less; in particular, there is less vertically downward component of the drag while the ball rises. Thus, it will reach a greater height and spend longer in the air.

It is also obviously wrong to take the horizontal component of velocity as being simply and uniformly increased by the velocity of the wind over the whole trajectory. Rather, the initial horizontal velocity is the same regardless of the wind, but it reduces less rapidly with the following wind.

 

[brotherbobby]

I assumed the cannon fires at at fixed but unknown angle? The same angle both days?

Yes.

 

[PeroK]

Yes.

Ah, I see what you've done. Introducing a new angle $\theta'$ - was it worth it? You're dealing with velocity components in any case, so there's no need to calculate the new angle. It doesn't give you anything you don't already know.