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I am currently doing a maths exploration but I cannot for the life of me figure out how to calculate the range of a projectile with a speed determined by the function (where t is time): $$ v = \frac {t+400}{73}-240$$

The angle of velocity is constantly changing and can be modelled by the function: $$\theta =e^{\frac{-t+97}{24.8}}+47$$

Would I need to calculate the angular velocity and find the acceleration by differentiating the speed?

Sorry if this all seems like a basic question, but my teachers weren't able to help me. Thank you for any answers :)

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  • $\begingroup$ It would be helpful to provide some additional information and context of those equations, currently they are little more than dimensionless mumbo-jumbo. $\endgroup$ – Soba noodles Nov 9 '16 at 23:22
  • $\begingroup$ For example, at t=0, velocity is negative. What kind of projectile is this? Is it projected from some height? $\endgroup$ – npojo Jan 22 '18 at 20:16
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Sorry if this all seems like a basic question, but my teachers weren't able to help me. Thank you for any answers :)

That's almost certainly because this problem is mathematically more demanding than you probably imagine.

I'm assuming $\theta$ to be the trajectory's angle to the horizontal.

Write the velocity $v$ as: $$v=v(t)sin\theta(t)$$ Where: $$v(t)=\frac {t+400}{73}-240$$ $$\theta(t)=e^{\frac{-t+97}{24.8}}+47$$ Now split the velocity into two components, $v_y(t)$ (vertical) and $v_x(t)$ (horizontal). As per Galileo these movements are completely independent of each other: $$v_y=v\sin\theta$$ With $v_y=\frac{dy}{dt}$: $$\implies y(t)=\int_0^tv_ydt$$ $$y(t)=\int_0^t\Big(\frac {t+400}{73}-240\Big)\Big(e^{\frac{-t+97}{24.8}}+47\Big)\sin\Big(e^{\frac{-t+97}{24.8}}+47\Big) dt\tag{1}$$ Once this expression $(1)$ is found, use it to determine how long ($t_{ground}$) it takes for the projectile to reach the horizontal again: $y(t_{ground})=0$.

Similarly for the horizontal displacement, we get: $$v_x=v\cos\theta$$ $$x(t)=\int_0^tv_xdt$$ $$x(t)=\int_0^t\Big(\frac {t+400}{73}-240\Big)\Big(e^{\frac{-t+97}{24.8}}+47\Big)\cos\Big(e^{\frac{-t+97}{24.8}}+47\Big) dt\tag{2}$$ Inserting $t_{ground}$ into $(2)$ then gives you the desired range.

The problem is that neither integrations $(1)$ or $(2)$ are particularly straightforward or easy to carry out. If you do want to pursue this, I suggest the following (unverified) strategy.

Substitute as follows:

$$\theta(t)=e^{\frac{-t+97}{24.8}}+47\tag{3}$$

Extract $t$ from $(3)$, then compute:

$$dt$$

and:

$$v(t)$$

The integrals should become of the form:

$$y(\theta)=\int f(\theta)\sin\theta d\theta$$

And should be manageable.

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  • $\begingroup$ If $\theta$ was the angle to the horizontal, shouldn't it be able to reach zero ? (It seems that $\theta\geq 47$) $\endgroup$ – user1583209 Dec 8 '17 at 21:52
  • $\begingroup$ No, theta is the angle the velocity vector makes with the horizontal, so it only ever increases. $\endgroup$ – Señor O May 4 '18 at 22:16

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