Question

A ball is kicked off of a roof at 23 m/s [R 25° U]. What is the height of

Answer 1

Answer:

Explanation:

Considering the fact that we ave been given an angle of inclination here, we best use it! That means that the velocity of 23 m/s is actually NOT the velocity we need; I tell my students that it is a "blanket" velocity but is not accurate in either the x or the y dimension of parabolic motion. In order to find the actual velocity in the dimension in which we are working, which is the y-dimension, we use the formula:

$v_{0y}=v_0sin\theta$ and filling in:

$v_{0y}=23sin(25)$ which gives us an upwards velocity of 9.7 m/s. So here's what we have to work with in its entirety:

$v_{0y}=9.7m/s$

a = -9.8 m/s/s

t = 2.8 seconds

Δx = ?? m

The one-dimensional motion equation that utilizes all of these variables is

Δx = $v_0t+\frac{1}{2}at^2$ and filling in:

Δx = $9.7(2.8)+\frac{1}{2}(-9.8)(2.8)^2$ I am going to do the math according to the correct rules of significant digits, so to the left of the + sign and to 2 sig fig, we have

Δx = 27 + $\frac{1}{2}(-9.8)(2.8)^2$ and then to the right of the + sign and to 2 significant digits we have

Δx = 27 - 38 so

Δx = -11 meters. Now, we all know that distance is not a negative value, but what this negative number tells us is that the ball fell 11 meters BELOW the point from which it was kicked, which is the same thing as being kicked from a building that is 11 meters high.