If an object is dropped, how fast will it be moving after falling #25 m#?

Question

If an object is dropped, how fast will it be moving after falling #25  m#?

William Audy · Accepted Answer

Assuming no air resistance, it will be moving at #v=sqrt(2ad) = sqrt(2*9.8*25)=sqrt(490)=22.1# #ms^-1#

The following formula can be used to calculate how quickly an object will accelerate after traveling a given distance:

#v^2=u^2+2ad#

Where:

#v# = final velocity #(ms^-1)# #u# = initial velocity #(ms^-1)# #a# = acceleration #(ms^-2)# #d# = distance #(m)#

In this case, we assume that the initial velocity is #0#: the question says the object was 'dropped', not 'thrown'.

We can leave out the #u# term and take the square root of both sides to make #v# the subject:

#v=sqrt(2ad)#

Entering our values and incorporating the gravitational acceleration:

#v = sqrt(2*9.8*25)=sqrt(490)=22.1# #ms^-1#

Because air resistance is much weaker than the force of gravity, we have assumed that it either doesn't exist or that it can be ignored.

Leo Abeyta · Answer

undefined The final velocity of a freely falling object after falling 25 meters can be calculated using the formula for uniformly accelerated motion. Assuming no air resistance, the final velocity (v) can be found using the equation:

$$v^2 = u^2 + 2as$$

Where:
- $v$ is the final velocity,
- $u$ is the initial velocity (which is 0 for an object dropped),
- $a$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$),
- $s$ is the displacement (in this case, 25 meters).

Solving for $v$:

$$v = \sqrt{2as}$$