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Let's assume a ball is thrown upwards with an initial velocity V and gravitational acceleration -A is acting on it in downward direction At some height H the particle with reach its max height and start falling down in negative direction I want to calculate the distance travelled by the ball and not the displacement

I can use s = ut + ½at² to calculate displacement but what i want is the distance travelled

For example, when the ball reaches the ground the displacement will be 0 but distance travelled will be twice the max height reached

How can I calculate distance travelled at any time t?

Omkar T
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Mar 09 '23 at 18:40
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    I believe it's pretty clear – Omkar T Mar 09 '23 at 18:44

3 Answers3

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Let's make this a bit more general. Consider $s:[0,T]\rightarrow \mathbb{R}^n$, a parameterization of the trajectory the ball follows ($s(t)=$ position of the ball at time $t$, and $n=1$ in your case).

In a small fraction of time $dt$, your ball moves along the trajectory, approximately in a straight line tangent to the trajectory, by an amount $|g'(t)|dt$. Sum these amounts for all times to get your desired length: $$\int_0^T |s'(t)|dt$$

Lilla
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  • Thanks for the response but I'd appreciate a simplified solution not dealing with integration, being a software engineer – Omkar T Mar 09 '23 at 18:55
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Consider a short time $\delta t$. The ball's velocity at this point is $v_0 - a t$, where $v_0 = v (t = 0)$, can be determined from simple kinematics. The distance travelled in this time is $\left| v_0 - a t \right| \delta t$. To find the distance traveled from $t = 0$ to some arbitrary $t$, simply integrate

$$d = \int_0^t \left| v_0 - a t' \right| d t'$$

There is no clean closed form for this. If we take $a$ and $v_0$ to be positive (downward acceleration, upwards initial velocity), then for $t \le \frac{v_0}{a}$ which corresponds to the upwards part of the motion, the distance will be identical to the displacement: $d (t) = v_0 t - \frac{1}{2} a t^2$. At $t_c = \frac{v_0}{a}$, the distance traveled is $d_c = \frac{v_0^2}{2 a}$. For $t > \frac{v_0}{a}$, the distance is $d_c$ plus the magnitude of the downward displacement. In particular, $d (t) = d_c + \frac{1}{2} a (t - t_c)^2$.

K. Jiang
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  • Sorry to disappoint you but can you care to explain what is t and tc in last equation? – Omkar T Mar 09 '23 at 19:05
  • $t$ is your time coordinate. $t_c = \frac{v_0}{a}$ was defined - it's the time at which the projectile reverses direction. – K. Jiang Mar 09 '23 at 19:07
  • As per my little understanding, shouldn't it be -0.5a(t-tc)^2 as acceleration is negative? and if so how can I prove 0.5a(t-tc)^2 == dc (so d(t) = 0 at t =.. well t) – Omkar T Mar 09 '23 at 19:25
  • "shouldn't it be -0.5a(t-tc)^2": No, because you wanted the distance, not displacement. Once the ball starts moving down, the distance traveled continues to increase (we don't care about the direction of motion). I'm not quite sure what you mean by your second question, but note that $d (t_c) = d_c + \frac{1}{2} a (t_c - t_c)^2 = d_c$ by definition. – K. Jiang Mar 09 '23 at 19:28
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    Understood, Had a misunderstanding there. Thanks for your time :) – Omkar T Mar 09 '23 at 19:46
  • Sure, glad to help! – K. Jiang Mar 09 '23 at 19:47
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If you want a simple formula not involving integration, then try the following: First note that the particle reaches its maximum height $H$ where $$H=\frac{V^2}{2A}$$ at time $$T=\frac{V}{A}$$

(Taking the positive value of $A$, e.g. $9.8 ms^{-2}$)

So, for $t\leq T$, the distance travelled is $$s=Vt-\frac12At^2$$

For $t>T$, the total distance is $$2H-s$$ $$=\frac{V^2}{A}-Vt+\frac12At^2$$

Note that the second formula will still be valid when the particle descends to a level below its initial position.

I hope this helps.

David Quinn
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