0
$\begingroup$

Two balls with exactly the same size and shape, but different mass, are launched at the same velocity 90 degrees to a flat plane. When air resistance is considered, the object with the larger mass will have a:

Longer/shorter flight length with a greater/lesser maximum height.

My logic:

The heavy ball reaches a greater maximum height because on its way up there less acceleration in the downward direction.

Force of drag is proportional to velocity. They both have the same starting velocity so they both experience the same initial drag force. However, the heavier object experiences a lesser acceleration due to that force.

Now, the direction of that force will be directly down before the ball reaches its peak because drag opposes the direction of the velocity.

Acceleration = $10ms{^-2}$ (gravity) + $a_{drag}$

$$a_{large} = -10 - \frac{F_{drag}}{m_{large}}$$

$$a_{small} = -10 - \frac{F_{drag}}{m_{small}}$$

The part I don't get is the longer time in the air. The more massive object accelerates down faster than the less massive object after reaching the maximum height.

$\endgroup$
  • 1
    $\begingroup$ Related brain teaser. Consider a 100 km drive. Which is faster, driving the whole distance at 100 kph or driving halfway at 90 kph and halfway at 110 kph? Your situation is more complicated, of course, because the massive one travels further and it is acceleration, not velocity that is varied, but there are some similar schemes for arguing it. $\endgroup$ – dmckee Jun 14 '16 at 2:29
1
$\begingroup$

The last sentence is partly wrong - the heavier object does not accelerate faster when coming down; g is identical for both.

The drag however, again breakes the lighter object more than the heavier object; but the lighter object start its fall much lower. At the end the winner depends on the parameters.

  1. Imagine throwing a steel ball and a ping-pong ball of same size really hard. The ping pong ball will fall down quickly, and hit earth long before the steel ball is done climbing; so the steel ball flies longer.
  2. Now imagine throwing them rather lightly up, but from a cliff top. The steel ball will quickly turn and fall with high speed, whereas the ping pong ball will start to fall earlier, but take its time to come down, so it will fly longer.
$\endgroup$
  • 2
    $\begingroup$ g may be identical for both; but the net acceleration is not going to be g. The heavier mass would accelerate faster when moving down. $\endgroup$ – JMac Sep 7 '17 at 14:26
0
$\begingroup$

Imagine throwing a steel ball and a ping-pong ball of same size really hard. The ping pong ball will fall down quickly, and hit earth long before the steel ball is done climbing; so the steel ball flies longer.

Now imagine throwing them rather lightly up, but from a cliff top. The steel ball will quickly turn and fall with high speed, whereas the ping pong ball will start to fall earlier, but take its time to come down, so it will fly longer.

Keep in mind that there is a difference between force and acceleration...

In order to launch a ball at the mass of 10 kg at 20 m/s^2, you would need:

F = ma
  = 10*20
  = 200 N

whereas a ball at the mass of 1 kg at 20 m/s/s would need 20 N of force


Anyway, a heavier ball would most definitely fall faster. It is thanks to intertia. Here is the equation to drag:

Fd = -.5*rho*|V|*V*Cd*A

where,

rho is the air density

V is the velocity vector

|V| is the absolute value of the velocity vector

Cd is the drag coefficient (0.47 for spheres)

A is the cross sectional area (radius of sphere^2*pi)

So if Fd is 5N, and because gravity is constant, an object of 5 kg would have 1 m/s^2 of drag:

a = f/m
a = 5/5
a = 1 m/s^2

where an object of 0.5 kg would have 10 m/s^2 of drag:

a = F/m
a = 5/.5
a = 10 m/s^2
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.