# College Physics 2.52 – Object dropped from a given height

## 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.