6.2 CHAPTER 6. MOTIONIN TWO DIMENSIONS
t = 0, 5 s
Δx = 1, 225 m
Step 4 : Find the height and thetime taken for the second motion.
For the second part of the motion we
have:
- vi= 0 m· s−^1
- Δx =−(20 + 1,225) m
- g =− 9 ,8 m· s−^2
Therefore we can use Δx = vit+^12 gt^2
to solve for the time.
Δx = vit +
1
2
gt^2
−(20 + 1,225) = (0)× t +
1
2
× (− 9 ,8)× t^2
− 21 ,225 = 0− 4 , 9 t^2
t^2 = 4, 33163...
t = 2, 08125... s
20 m
g =− 9 , 8 m·s−^2
vi= 0 m·s−^1
Δx =− 21 , 225 m
Step 5 : Graph of position vs. time
The ball starts from a position of 20 m (at t = 0 s) from the ground
and moves upwards until it reaches (20 + 1,225) m (at t = 0,5 s).
It then falls back to 20 m(at t = 0,5 + 0,5 = 1,0s) and then falls
to the ground, Δ x = 0 m at (t = 0,5 + 2,08 = 2,58 s).
