Problem 2-95 Linear Kinematics - Part 5 - (a)

A camera is set up to take photographs of a ball undergoing vertical motion. The camera is \(5.2 \;m\) above the ball launcher, which can launch the ball with an initial velocity of \(17 \;m/s\) upward. Assuming that the ball goes straight up and straight down past the camera, at what times will the ball pass the camera?

Accumulated Solution


\(y = v_{0^t} + (1/2)at^2 \\ 5.1 = 17t - 4.9t^2 \\ y = \frac{-b}{2a} \pm \frac{\sqrt {b^2 - 4ac}}{2a} \\ t = \frac{3.47}{2} \pm \frac{\sqrt{12.041 -4.245}}{2} \\ = 1.735 \pm 1.396 \\ = 3.1\; s \; \text{or} \; 0.34 \; s\)

No.

It is a common mistake for students to believe that one root is "right" and the other "wrong". This is probably because many questions only ask for one answer ignoring another possibility. Both roots, however, have physical meaning as in this question. The ball passes the camera \(0.34\; s\) after launch on the way up and at \(3.1\; s\) on the way down.

You have completed this problem.