Question

An object is thrown vertically into the air. Considering air resistance, the object's time coming down compared to its going up is______?
less
more
zero
the same

Answer 1

Answer:

Its answer is more as when ball is going upward, air resistance supports the motion but when ball comes down, air resistance opposes the motion.

Answer 2

Answer:

The object's time coming down compared to its going up is more.

Explanation:

Here,

The acceleration due to gravity is denoted by g.

Let the retardation due to the air drag is denoted by a.

The speed at the highest point is denoted by v.

Let the speed of the object thrown vertically into the air be u.

The time of the upward motion is denoted by $t_{u}$.

The time of the downward motion is denoted by $t_{d}$.

The net acceleration of the object for upward motion is denoted by $a_{u}$.

The net acceleration of the object for downward motion is denoted by $a_{d}$.

Now,

v = 0m/s

u and g are constant.

Now,

For the upward motion,

$a_{u}$ = g + a

Then,

By the equation,

$v=u-a_{u}t_{u} \\0=u-(g+a)t_{u} \\t_{u} = \frac{u}{g+a}$       .......(1)

Now,

For the downward motion,

$a_{d}$ = g - a

Then,

By the equation,

$v=u-a_{d}t_{d} \\-u=0-(g-a)t_{d} \\t_{d} = \frac{u}{g-a}$...........(2)

By comparing equations 1 and 2,

$t_{d}$ > $t_{u}$                 ∴ The lesser the denominator higher the time.

So, the object's time coming down compared to its going up is more.