Question

I have a question from example 2-20

Answer 1

@dan815
As it shows \(y = y_0 +v_0t+\dfrac{1}{2}at^2\) and they set y = -50. they calculate the time the ball hits the ground with the total distance the ball travel is 50 m. I don't understand why they don't count the distance from the cliff to the highest point where v =0??

Answer 2

We don't count so because the acceleration is uniform through out the motion so you don't need to worry about that :) A single motion equation will yield the same result.

Answer 3

for part b) they said that the total distance the ball travel is 73.0 m. why it's not the y for the calculation above? why it's just -50?

Answer 4

The total distance is the distance up plus the distance down, but that is not imediately irrelevant to the equation they use.

They use \(y=y_0+v_0\ t+\frac{1}{2}a\ t^2\). It is true in all scenarios with constant acceleration.

There, the \(y\) values are position: vertical position in this problem.

So they set the top of the cliff to be at \(y=0\). That is the initial position, which is \(y_0\) in the formula, so \(y_0=0\).

Then the cliff bottom is the final position, which is \(y\). That is 50 below the top of the cliff, so it is 50 below 0, which is -50. So \(y=-50\).

The final position is a function of time, so time is your domain. If you don't restrict your domain, then you'll see the results of the equation that don't matter, because the equation does not hold true for the entire domain. That's like when the ball hits the ground - no more constant acceleration. The calculation will be way off after that.

Answer 5

You have the top of the cliff to be \(y_0\), and the moment in time of throwing the ball to be \(t=0\).