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a) $v_x=10m/s$
b) $h=2.2m$

Giancoli 6th Edition, Chapter 6, Problem 44

(2:21)

Chapter 6, Problem 44 is solved.

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Transcript for this Giancoli solution

We know that the 'y' component of the velocity of this ball will be 0, at its maximum height. So at the top of its arc, its final speed is going to be the 'x' component of its velocity. This part isn't going to change, because there's no friction. And that is 'v' times, cos theta. So that's 12 meters per second, times cosine 33 degrees, which is 10.06. Which we'll say is 10 meters per second. So that's for part a.

In part b, we'll say that, we're going to find the maximum height of it now. And the initial kinetic post potential is equal to the final kinetic post potential. Initially we'll use its initial launch position as the reference point, so we'll say it has no height at all. And then we can write this one half 'mv' initial squared is, one half 'mv' final squared, plus 'mgh', this is the gravitational potential at the top of the path, the 'm's are a common factors which cancel. We're going to solve for 'h' here, so we'll subtract this from both sides, and also divide by 'g'. So we have 'h' equals, 'vi' squared minus 'vf' squared divided by 2'g', and substituting in numbers, 'h' is 12 meters per second, initially. Then its final speed is 10.06 meters per second, square that, divided by 2, times 9.8. We get a height of 2.2 meters.