2.4.2 Vertical Throw with Air Resistance

If you toss a tennis ball vertically upward, to the naked eye it probably looks like it takes the same amount of time to go up and come down. But if you toss a balloon instead, it is pretty obvious that it takes a lot more time to come down than go up. What causes this asymmetry?

Roughly speaking, the balloon goes through the following 3 stages:

During the ascent, both the air resistance R and gravitational pull mg act downward, resulting in a (downward) net force larger than mg, and (downward) acceleration larger than 9.81 m s^-2.

$\begin{aligned}{{F}_{{net}}}&=mg+R\text{ }\\a&=(mg+R)\div m=g+\frac{R}{m}\text{ }\Rightarrow \text{ }a>g\end{aligned}$

As the balloon slows down during its ascent, R decreases, causing the acceleration to decrease towards 9.81 m s^-2.

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When the balloon comes to an instantaneous rest at the top of its motion, air resistance is zero. The balloon experiences only mg. So its acceleration is $mg\div m=g$ . This is the only instant when the acceleration is 9.81 m s^-2.

$\begin{aligned}{{F}_{{net}}}&=mg\\a&=mg\div m\text{ }\Rightarrow \text{ }a&=g\end{aligned}$

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During the descent, R acts upward while mg acts downward, resulting in a (downward) net force smaller than mg, and (downward) acceleration smaller than 9.81 m s^-2.

$\begin{aligned}{{F}_{{net}}}&=mg-R\text{ }\\a&=(mg-R)\div m&=g-\frac{R}{m}\text{ }\Rightarrow \text{ }a<g\end{aligned}$

As the balloon speeds up during its descent, R increases, causing the acceleration to decrease towards 0 m s⁻².

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Air resistance always acts in opposite direction to velocity. The air resistance acting in different directions during the ascent and descent is the cause of the asymmetry in the balloon’s motion!

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v-t graph (upward is positive convention)

The v-t graph starts from the launch speed v_i, decreases (as the balloon rises), reaches zero (when the balloon is at the peak), then increases in the negative direction (as the balloon falls), and reaches the final speed v_f (when the balloon returns to the original position).
The graph is a smooth curve, with gradient becoming gentler and gentler. At v=0, when the balloon is at the peak, the gradient of the v-t graph should be 9.81 m s^-2.
The green and blue areas are equal since they represent the upward distance and downward distance travelled by the balloon.
Because of the curvature of the graph, it is clear that the landing speed v_f is always going to be smaller than the launch speed v_i, and that it always takes a longer time for the balloon to come down than go up. Physically, this is because the balloon decelerates much faster (on the way up) than it accelerates (on the way down). The upward motion has a higher average speed than the downward motion.

a-t graph

The a-t graph is negative throughout.
The magnitude of the (downward) acceleration decreases continuously. It is more than 9.81 m s⁻² on the way up, but less than 9.81 m s^-2 on the way down. At the instant when the balloon is at the peak, the acceleration is exactly 9.81 m s^-2.
The green area ought to be larger than the blue since ${{v}_{i}}>{{v}_{f}}$ .

s-t graph:

Because of air resistance, the s-t graph is not a symmetrical quadratic curve.
Since gradient of s-t graph represents the velocity, the graph has a steeper gradient=v_i at $t=0$ , and a gentler gradient=v_f at the end.