Question

Assume you drop a package from the third floor and it takes 3.60 seconds to reach the floor. If the object was In complete free fall what would the velocity be right before hitting the floor ?

Answer 1

Answer:

The velocity of the package before hitting the floor is 35.28 m/s.

Explanation:

Given:

The package is dropped, so initial velocity is, $u=0\ m/s$

Time taken to reach the floor is, $t=3.60\ s$

The motion is a complete free fall motion.

When the body is in free fall, the only acceleration acting on the body is due to gravity.

Therefore, the acceleration of the package is, $a=g=9.8\ m/s^2$

Now, final velocity of the package is, $v=?$

In order to find the final velocity, we need to use the Newton's equation of motion that relates initial velocity, final velocity, acceleration and time taken.

So, the equation of motion used is given as:

$v=u+at$

Plug in the given values and solve for 'v'. This gives,

$v=0+9.8\times 3.60\\v=35.28\ m/s$

Therefore, the velocity of the package before hitting the floor is 35.28 m/s.

Answer 2

Answer:

35.316 $m/s$

Explanation:

Given: $t=3.60\ sec$

In free fall velocity does not depend upon height of fall.

Now, we will use the relation between final velocity (v) and initial velocity (u) and time (t) under free fall.

That is $v=u+gt$

Where 'g' is acceleration due to gravity which is $9.81 m/s^2$

Here initial velocity (u) is zero as package was dropped from rest.

Plugging the values in the equation, we get.

$v=u+gt\\\\v=0+9.81(t)\\\\v=9.81(3.60)\\\\v=35.316\ m/s$

Hence, the velocity with which the ball will hit the ground will be $v=35.316\ m/s$.