Question #a06f3
1 Answer
Here's what I got.
Explanation:
The first thing to do here is to calculate the maximum height that the stone reaches.
At the top of its trajectory, i.e. at maximum height, its velocity will be equal to zero, which means that you can write
#0 = v_0^2 - 2 * g * h_"max"#
Here
Rearrange to solve for
#v_0^2 = 2 * g * h_"max" implies h_"max" = v_0^2/(2 * g)#
Plug in your values to find
#h_"max" = (20^2 "m"^color(red)(cancel(color(black)(2))) color(red)(cancel(color(black)("s"^(-2)))))/(2 * 9.81 color(red)(cancel(color(black)("m")))color(red)(cancel(color(black)("s"^(-2))))) = "20.4 m"#
Now, you know that the stone was caught on its way down at a height of
This essentially means that the stone traveled a total of
#h_"down" = "20.4 m" - "5 m" = "15.4 m"#
on its way down from its maximum height. Since its velocity was equal to zero at
#v_"caught"^2 = 0^2 + 2 * g * h_"down"#
Plug in your values to find
#v_"caught" = sqrt(2 * "9.81 m s"^(-2) * "15.4 m") = color(darkgreen)(ul(color(black)("17.4 m s"^(-1))#
To find the time of the trip, we can use the fact that the stone is passing through the point at which it gets caught, i.e. through
You can thus say that you have
#h_"down" = v_0 * t^2 - 1/2 * g * t^2#
Plug in your values and rearrange the equation as
#4.905 * t^2 - 20 * t^2 + 5 = 0#
This quadratic equation will produce two values for
#{(t_1 = "3.81 s"" "color(darkgreen)(sqrt())), (t_2 = color(red)(cancel(color(black)("0.268 s")))) :}#
Since we're looking for the time needed for the stone to pass through
Therefore, the trip took a total of
I'll leave both answers rounded to three sig figs, but keep in mind that you only have one sig fig for the initial velocity and height at which the stone is caught.