A boy throws a ball straight upward and the ball returns to the boy's hands after a time of 3 s. How fast was the ball thrown & how high did it rise?
Answers
It takes 3s for the whole trip.
Since acceleration/gravity is constant. The
upward or rising part of the trip is symmetrical with the falling part.
That means I can divide the trip into two equal parts. So the upward trip will take 1.5 seconds and the falling trip likewise will take 1.5 seconds.
Acceleration of gravity is about 10(m/s)/s in the downward direction. It lowers the upward velocity by 10(m/s)/s every second.
The upward trip ends when upward velocity reaches 0
If it took 1.5 seconds for upward trip to end then Om/s=Xm/s - 1.5s(10m/s^2)
X is the initial velocity and = 15m/s
Let's do avg velocity (THIS IS VERY TRICKY AND
MESSES STUDENTS UP)
Avg velocity of the whole trip (up and down) is 0
No calculation required. Know why? Because velocity is a VECTOR that means direction
matters and velocity is defined as DISPLACEMENT(vector) over time.
Do you remember how displacement is the difference between the INITIAL and FINAL positions (the path it takes doesn't matter).
So where do we start and where do we end? We start AND end in the hand of boy. We end in the same place we began (path DOES NOT MATTER) That means displacement is 0 even though the ball traveled some distance up and some distance down.
So avg velocity depends on initial and final
position
Avg speed is SCALAR and doesn't care about direction. So we have to calculate it.
We can take an average of speed by taking the initial and final velocity and dividing by two. Since upward and downward trip are symmetrical (the same) we only need to look at one of the parts.
Remember that is upward trip initial velocity is 15m/s and final velocity is Om/s (symmetrically therefore the downward trip has an initial velocity of 0 and final velocity of 15m/s(when the ball returns to the boy)
Take an avg speed:
(15m/s + 0m/s)/2