- #1

TheRedDevil18

- 408

- 1

## Homework Statement

**Please check my work.**

Two students are on a balcony 19.6m above the ground. One guy throws a ball vertically downwards at 14.7m/s; at the same time the other guy throws a second ball vertically upwards at the same speed. The second ball just misses the balcony on the way down.

2.3.1) What is the difference in the balls flight time?

2.3.2) What is the velocity of each ball as they strike the ground?

2.3.3) How far apart are the balls 0.8s after they are thrown?

## Homework Equations

vf^2 = vi^2 + 2a*y

delta y = vi*t + 1/2*a*t^2

delta y = vf+vi/2 * t

vf = vi + a*t

## The Attempt at a Solution

2.3.1) For the ball thrown upwards I decided to first find the maximum height it reached. I used the first equation from the list and got an answer of 11.03m. I then used the third equation to find the time and got 1.5s (another 1.5s for coming down=3s). I then worked out the time for the ball thrown downwards to be 1s. My final answer for the difference in flight time is 4-1=3s.

2.3.2) Each ball would strike the ground at the same speed.

vf = vi+a*t

= 14.7+9.8*1

= 144.06m/s

2.3.3) Using the heights of the two, I worked the distance for the first ball to be 8.624m and the second ball to be 14.896m. Subtracting the two would give a distance of 6.272m as my final answer.

All I am asking is for someone to please check my work because I am not sure if I have approached this problem in the correct way.