# Falling Objects| Kinematics| Dropped Object| College Physics| Problem 2.52

### An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

Solution:

PART A

We can calculate for the distance traveled in 1 second using the formula $y=v_ot-\frac{1}{2}gt^2$

Substitute the given values $y=0-\frac{1}{2}\left(9.80\:m/s^2\right)\left(1\:s\right)^2=-4.9\:m$

The object has traveled 4.9 meters downward during the first 1 second.

PART B

We can use the formula $v^2=\left(v_0\right)^2-2gy$

Substitute the given values $v^2=\left(0\right)^2-2\left(9.80\:m/s^2\right)\left(-75.0\:m\right)$ $v^2=1470$ $v=\sqrt{1470}=38.34\:m/s$

The object has a velocity of 38.34 m/s when it hits the ground.

PART C

We need to find the total time of travel of the object, using the formula $y=v_0t-\frac{1}{2}gt^2$

Substitute the given values $-75.0\:m=0-\frac{1}{2}\left(9.80\:m/s\right)t^2$

Solve for time, t $t=\sqrt{\frac{2\left(75.0\:m\right)}{9.80\:m/s^2}}=3.91\:s$

Therefore, the object has traveled 75.0 meters during the entire flight of 3.91 seconds.

Next, we determine the total distance traveled during the first 2.91 seconds of the flight. $y=v_0t-\frac{1}{2}gt^2$

Substitute the given values $y=0-\frac{1}{2}\left(9.8\:m/s^2\right)\left(2.91\:s\right)^2=-41.49\:m$

Therefore, the object has traveled 41.49 meters downward during the first 2.91 seconds. Between the time 2.91 and 3.91 seconds, we have the last 1 second of the flight. The distance traveled for this second is $y=75.0\:m-41.49\:m=33.51\:m$

The object has traveled 33.51 meters for the last second of the flight.