You are here

a)$27 \textrm{m/s}$
b)$37 \textrm{m}$
c)$1.4 \textrm{s ago}$
d)$4.1 \textrm{s}$

Giancoli 6th Edition, Chapter 2, Problem 48


Chapter 2, Problem 48 is solved.

View sample solution


In problem B why is the velocity initial in the equation the 13m/s instead of the 26m/s found in problem A? And then going into problem c the velocity initial is 26m/s again, and velocity final is 13m/s instead of 0?

Hi jtorres89,

I can definitely see why this is confusing, so thank you for your question.

The difference between parts A, B, and C, has to do with perspective, and avoiding errors in one part spoiling the next part.

In part B, I chose to solve for the distance above the bottom window sill instead of above the ground, and then added the 28m position of the window sill above the ground to get the final answer of the ball height above the ground. The reason I did it this way was to remind students to, where possible, use numbers directly from the question, instead of answers you've calculated, just in case the calculation for part A is wrong so that it won't spoil part B. We could have used 27m/s in part B if we wanted, but this takes a small risk that if it's wrong, then we'll get part B wrong as well. By choosing the reference point in the question to be the bottom window, that means d=0 there, and t=0 there as well, and that the speed there, 13m/s, is the initial speed.

In part C the point of reference is the ground, instead of the window sill. The ground level is d=0, t=0, and the speed there is the initial speed. The initial speed at the ground is 27m/s, as calculated in part A (and sometimes I use calculated values in subsequent parts too). The final point is the bottom of the window, since we want to know the time it took to get from the ground to the window sill.

Hope this helps!