Problem 2-95 Linear Kinematics - Part 3 - (c)

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

diagram of camera capturing a ball moving vertically
\(y = v_{0^t} + (1/2)at^2 \\ 5.1 = 17t - 4.9t^2\)

Correct!
\(5.1 = 17t - 4.9t^2\)
or
\(t^2 - 3.47t = 1.0612 = 0\)

The solution of the quadratic equation
\(ax^2 + bx + c = 0\)  is:

(A)   \(y = \frac{b}{2a}\pm \frac{\sqrt {b^2 - 4ac}}{2ac}\)

(B)   \(y = \frac{-b}{2a} \pm \frac{\sqrt {b^2 - 4ac}}{2a}\)

(C)   \(y = \frac{-b}{2a} \pm \frac{\sqrt{b^2 + 4ac}}{2a}\)