Question

A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 6.00 m/s. Air resistance may be ignored, so the water balloon is in free fall after it leaves the thrower's hand. What is its speed after falling for 2.00 s?

Answer 1

Answer: 13.6 m/s

Step-by-step explanation:

Since we are told the water balloon is in free fall and the air resistance is neglected, we are dealing with constant acceleration (due gravity)
`g`. Hence, we can use the following formula to find the velocity
`V` of this balloon at a given time and then its speed:

`V=V_(o)+gt` (1)

Where:

`V_(o)=6 m/s` is the water balloon's initial velocity

`g=-9.8 m/s^(2)` is the acceleration due gravity (always directed downwards)

`t=2 s` is the time

Rewritting (1) with the known values:

`V=6 m/s+(-9.8 m/s^(2))(2s)` (2)

[tex]V=-13.6 m/s/tex] (3) This is the velocity of the balloon and the negative sign indicates is directed downwards

However, we were asked to find the speed, which is the magnitude of the velocity vector and is always positive.

Therefore, the speed of the balloon is 13.6 m/s