6.2 CHAPTER 6. MOTIONIN TWO DIMENSIONS
Step 4 : Find the maximum height (Δx) of the ball.
To find the maximum height we look at the initial motion of the
ball. We have the following:
t = 0, 2 s
g =− 9 ,8 m· s−^2
vf = 0 m· s−^1 (because the ball stops)
vi = +1,96 m· s−^1 (calculated above)
To calculate the displacement from the windowto the maximum
height (Δx) we use:
Δx = vit +
1
2
gt^2
Δx = (1,96)(0,2) +
1
2
(− 9 ,8)(0,2)^2
Δx = 0, 196 m
The maximum height ofthe ball is (4 + 0,196) =4,196 m above
the ground.
Step 5 : Find the final velocity(vf) of the ball.
To find the final velocityof the ball we look at the second part of
the motion. For this wehave:
Δx =− 4 , 196 m(because upwards is positive)
g =− 9 ,8 m· s−^2
vi = 0 m· s−^1
We can use (vf)^2 = (vi)^2 + 2gΔx to calculate the final velocity
of the ball.
(vf)^2 = (vi)^2 + 2gΔx
(vf)^2 = (0)^2 + 2(− 9 ,8)(− 4 ,196)
(vf)^2 = 82, 2416
vf = 9, 0687... m· s−^1
The final velocity of theball is 9,07 m·s−^1 downwards.
