CHAPTER 6. MOTIONIN TWO DIMENSIONS 6.2
time (s)
Position (m)
0,2 0,4?
1
2
3
4
5
SOLUTION
Step 1 : Find the height of the window.
The initial position of the ball will tell us how high the window
is. From the y-axis on the graph we can see that the ball is 4 m
from the ground.
The window is therefore4 m above the ground.
Step 2 : Find the time taken toreach the maximum height.
The maximum height iswhere the position-timegraph show the
maximum position - thetop of the curve. This iswhen t = 0,2 s.
It takes the ball 0,2s to reach the maximum height.
Step 3 : Find the initial velocity(vi) of the ball.
To find the initial velocity we only look at thefirst part of the
motion of the ball. That is from when the ballis released until it
reaches its maximum height. We have the following for this: In
this case, let’s choose upwards as positive.
t = 0, 2 s
g =− 9 ,8 m· s−^2
vf = 0 m· s−^1 (because the ball stops)
To calculate the initial velocity of the ball (vi), we use:
vf = vi+ gt
0 = vi+ (− 9 ,8)(0,2)
vi = 1,96 m· s−^1
The initial velocity of the ball is 1,96 m·s−^1 upwards.
