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I have an exercise that reads; You jump from a 10-meter-high tower, calculate the time needed and the velocity just before I touch the ground, regardless of the mass of the body falling.

I've already solved it, but I need help. In the class we learn about calculus, therefore I tried to use it in order to calculate the time and velocity. The thing I don't understand is why calculus and not using the first method written below?

My solution without calculus:

$$ g = v / t = 10 m/s^2$$ $$v = d / t$$ $$ => g = d / t^2 $$ $$ t^2 = d / g $$ $$ t^2 = 10/10 $$ $$ => t = 1 s $$ $$ v = d / t = 10 / 1 = 10 m/s $$

With calculus:

$$ v = Δd / Δt $$ $$ g = v / t => g = Δd / Δt * t $$ $$ Δd = g * t * Δt $$ $$ Δd = \int_{0}^{t} 10 × t × Δt => 10  \int_{0}^{t} t × Δt$$

$$10 × ½ t² => t = √2 $$

thanks in advance

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    By the way, you can use $\Rightarrow$ (\Rightarrow) or $\rightarrow$ (\rightarrow) to make implication signs. (The latter has an alias $\to$ (\to).) – 2017-02-26

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Your work doesn't really make sense to me the formula $d/t$ does not apply for changing velocity.

When something is falling off a tower, it's velocity changes of course.

It made sense to define velocity as displacement over time when something was moving at constant velocity. So we did.

But now something is not moving at constant velocity, what do we do.

We say that for a time period say $t$ in $t_f-\Delta t$ to $t_f$ if $\Delta t \to 0$ we approach constant velocity.

Surely in the matter of $.0001$ seconds the object cannot significantly change velocity, so then the definition of velocity for the constant case just about applies. But we can do better, we can let $\Delta t \to 0$. And that is when calculus comes in.

The quantity $d/t$ is only defined to be velocity when we have constant velocity, rather it is always defined as average velocity.

The same applies to acceleration, $\Delta v/t$ only a makes sense for the constant acceleration case, which we do have here.

We have then that $\lim_{t \to 0} \frac{\Delta v}{\Delta t}=a=-g$ from which we find $V=V_0+gt$ below:

For $\Delta t \to 0$:

$$\frac{\Delta v}{\Delta t}=-g$$

$$\Delta v=-g \Delta t$$

$$\int dv=\int -g dt$$

$$V=C-gt$$

Let $t=0$.

$$V=V_0-gt$$

This could have just been easily derived by $\Delta v/t=-g$ because here we have constant acceleration.

But because now we don't have constant velocity to derive the displacement formula we should use calculus, or in this case it is sufficient to use geometry. In any case as $g \approx 10$ we get,

$$t \approx \sqrt{2}$$

$$V \approx 0-g\sqrt{2}$$

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    Brilliant! One more question, I need to draw a graph to this thing, would the graph be a linear or exponential? To my mind, I think it would be exponential, right? – 2017-02-26
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    Glad it makes sense. The graph of velocity vs time is linear as seen by the equation $v(t)=v_0+gt$. The graph of displacement vs is quadratic as seen the equation in this case $d(t)=-\frac{1}{2}gt^2$ which comes from $\int v dt=\int-gt dt$ @Kodi – 2017-02-26