Animated Solution for Physics - Laws of Motion: A small ball of mass $m$ is thrown upward with velocity $u$ from the ground. The ball experiences a resistive force $mk{v}^2$, where $v$ is its speed. The maximum height attained by the ball is

Gravity Meets Drag

Imagine throwing a ball straight up into the air. In a perfect vacuum, the only force acting on it would be gravity, pulling it down with a constant force of $mg$. But we don't live in a vacuum! As the ball cuts through the air, it experiences a drag force.

The problem tells us this resistive force is proportional to the square of its speed, given by $mk{v}^2$. Because the ball is moving upwards, this drag force acts downwards, exactly in the same direction as gravity. This means the net force on the ball is the sum of these two downward forces.

According to Newton's Second Law, the net force equals mass times acceleration ($F_{net}=ma$). So, we can write:

Notice the negative signs? They are crucial! They indicate that both forces are acting in the negative $y$-direction (downwards), opposing the positive upward velocity. We can easily cancel the mass $m$ from both sides, leaving us with the acceleration:

Here is where the magic happens. We need to find the maximum height, which is a position ($y$). The standard definition of acceleration is $\frac{dv}{dt}$, but that introduces time into our equation. Instead, we use the powerful chain rule substitution $a=v\frac{dv}{dy}$. This directly links velocity and position!

Now we have a beautiful differential equation. To solve it, we must separate the variables. We gather all the $v$ terms on one side and the $y$ terms on the other:

It is time to integrate. We must set our limits carefully. The ball starts at the ground ($y=0$) with an initial velocity $u$. It reaches its maximum height ($y=H$) when its velocity drops to exactly $0$.

The right side is a trivial integration, yielding $-H$. The left side requires a classic $u$-substitution (or $t$-substitution in our case). Let $t=g+k{v}^2$. Differentiating this gives $dt=2kvdv$, which means $vdv=\frac{dt}{2k}$.

This transforms our integral into a simple logarithmic form:

Substituting our expression for $t$ back in and applying the limits from $u$ to $0$:

Plugging in the upper limit ($0$) gives $\ln (g)$. Plugging in the lower limit ($u$) gives $\ln (g+k{u}^2)$.

Using the property of logarithms that $\ln (A)-\ln (B)=\ln \left(\frac{A}{B}\right)$, we can combine the terms. We also multiply both sides by $-1$ to solve for $H$, which flips the fraction inside the logarithm:

Finally, dividing the terms inside the numerator by $g$ gives us our elegant final answer:

And there you have it! The maximum height is not just a simple quadratic relation anymore; it is governed by a logarithmic curve due to the relentless drag of the air.