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In a self exercise i was trying to solve this users question https://physics.stackexchange.com/questions/163822/what-is-the-initial-velocity-if-angle-and-range-is-given

My aim is 1) not to use the standard equations of motion but rather calculus; 2) to improve my ability to solve these type of questions.

The question states that the angle at which the projectile is launched is $45^\circ$. We know that $v_x=v_y$ which makes the calculations easier.

I started by $v_f=v_i+\int^{t_f}_{t_i}(a)dt$ thus $0=v_i-9.8\int^{t_f}_{t_i}dt$ and $v_i=9.8t$

$s_f=s_i+\int^{t_f}_{t_i}(v)dt$ in the y direction $s_f=0+\int^{t_f}_{t_i}(9.8t)dt$ in my first atempt i substituted $s_f$ for 12.5 since $v_x=v_y$ but this is obvoiusly wrong since $s_i$ will not reach 12.5m in $y$ displacement due to gravity, but what i know it in this interval the displacement in $x$ will be 12.5m

Any advise? to continue from here in relating the displacement in the $x$ axis to that in the $y$ axis

ps. I can see that the final answer should be given $v=\sqrt{v_x^2+v_y^2} 45^\circ$

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  • $\begingroup$ Why is the final velocity zero? If a ballistic rocket is launched, the place on which it falls is terribly damaged. So, the final velocity isn't zero. $\endgroup$
    – Sofia
    Feb 7, 2015 at 19:38
  • $\begingroup$ Final velocity in the $y$ must be zero as this is where the arrow reverses direction back to ground level $\endgroup$
    – Gobabis
    Feb 7, 2015 at 19:40

1 Answer 1

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You don't have to use the standard equations of motion, you can just deduce them from Newton's laws:

You have a force $\mathbf{F}=m\times -9.8 \: \mathbf{\hat{e}_y}$ acting on the arrow. Newton's second law is $\mathbf{F}=m\frac{d\mathbf{v}}{dt}$. Integrating for the $x$ and $y$ coordinates yield:

$$\frac{dv_y}{dt} = -9.8$$

$$\int_{v_{yi}}^{v_{y}}dv_y = \int_0^t-9.8dt $$

$${v_{y}}-{v_{yi}}=-9.8t$$

$$\frac{dv_y}{dt} = v_{yi}-9.8t$$

$$\int_{y_i}^{y}dv_x = \int_0^tv_{yi}-9.8dt $$

$$y=y_i+v_{yi}-4.9t^2$$

And for the $x$ coordinate:

$$\frac{dv_x}{dt} = 0$$

$${v_{x}}-{v_{xi}}= 0$$

$$x-x_i=v_{xi}t$$

These are the parametric equations of motion for projectiles in uniform gravity. Solve the first one for $t$, given the initial velocity and starting and ending on the ground $x_i=x=0$, to get the time it takes for the arrow to fall to the ground, then plug that into the second equation to get the initial velocity, given the distance traveled horizontally ($y-y_i$).

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