Alowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water.
(a) at what horizontal distance from the edge is the diver 0.800 s after pushing off?
(b) at what vertical distance above the surface of the water is the diver just then?
(c) at what horizontal distance from the edge does the diver strike the water?

(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:

Where is the horizontal distance at a time , is the initial horizontal position which in this case will be zero: . And is the initial speed:

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: so the horizontal distance is:

The answer for (a) is 1.6m

Now, to solve (b) we need the equation for vertical distance:

Where is the vertical distance at a time , is the initial vertical distance: And is the gravitational acceleration: }

(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:

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

if

and since

At the car hits the water, so the horizontal distance at that time is:

the answer to (c) is 2.86s

your answer is, orbit around the nucleus

answer; how far out of place which the object finished relative to the starting location;