Respuesta :
Answer:
(a) 1.6m
(b) 6.86m
(c) 2.86s
Explanation:
This is a horizontal shooting problem, so we will use the following equations.
For horizontal distance x we use:
[tex]x(t)=x_{0}+v_{0}t[/tex]
Where [tex]x(t)[/tex] is the horizontal distance at a time [tex]t[/tex], [tex]x_{0}[/tex] is the initial horizontal position which in this case will be zero: [tex]x_{0}=0[/tex]. And [tex]v_{0}[/tex] is the initial speed: [tex]v_{0}=2m/s[/tex]
So to solve (a):
(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?
we have that the time is: [tex]t=0.8s[/tex] so the horizontal distance is:
[tex]x(0.8)=(2m/s)(0.8s)1.6m[/tex]
The answer for (a) is 1.6m
Now, to solve (b) we need the equation for vertical distance:
[tex]y(t)=y_{0}-\frac{1}{2}gt^2[/tex]
Where [tex]y(t)[/tex] is the vertical distance at a time [tex]t[/tex], [tex]y_{0}[/tex] is the initial vertical distance: [tex]y_{0}=10m[/tex] And [tex]g[/tex] is the gravitational acceleration: [tex]g=9.81m/s^2[/tex] }
(b) At what vertical distance above the surface of the water is the diver just then?
The time is the same as in the last question: [tex]t=0.8s[/tex]
[tex]y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.8)^2[/tex]
[tex]y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.64s^2)[/tex]
[tex]y(0.8)=10m-\frac{1}{2}(6.2784m)[/tex]
[tex]y(0.8)=10m-(3.1392)[/tex]
[tex]y(0.8)=6.86m[/tex]
the answer to (b) is 6.86m
(c) At what horizontal distance from the edge does the diver strike the water?
To solve (c) we first need to know how long it took to reach the height of the water, that is, for what time y = 0
Using [tex]y(t)=y_{0}-\frac{1}{2}gt^2[/tex]
if [tex]y(t)=0[/tex]
[tex]0=10-\frac{1}{2}gt^2\\-10=-\frac{1}{2}gt^2\\20=gt^2\\\frac{20}{g}=t^2\\ \sqrt{\frac{20}{g}}=t[/tex]
and since [tex]g=9.81m/s^2[/tex]
[tex]t=\sqrt{\frac{20}{9.81} } =\sqrt{2.039}=1.43s[/tex]
At [tex]t=1.43s[/tex] the car hits the water, so the horizontal distance at that time is:
[tex]x(t)=x_{0}+v_{0}t[/tex]
[tex]x(1.43)=(2m/s)(1.43s)=2.86m[/tex]
the answer to (c) is 2.86s
