Question

A lowly 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?

Answer 1

To find the horizontal and vertical distances of the diver, we can use the equations of motion. After pushing off, the diver is 1.60 meters horizontally away from the edge after 0.800 seconds. The diver is 3.92 meters above the surface of the water at that time. The diver strikes the water 4.00 meters away from the edge.

Step-by-step explanation:

To solve this problem, we can use the equations of motion. Let's break down each part of the question:

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

To find the horizontal distance, we need to determine how far the diver travels horizontally in 0.800 seconds. We can use the formula:

d = v * t

where d is the distance, v is the initial horizontal velocity, and t is the time. In this case, v = 2.00 m/s and t = 0.800 s. Plugging in these values, we get:

d = 2.00 m/s * 0.800 s = 1.60 meters

Therefore, the diver is 1.60 meters away from the edge after 0.800 seconds.

(b) At what vertical distance above the surface of the water is the diver just then?

The vertical distance above the surface of the water can be found using the formula:

d = 1/2 * g * t^2

Where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. In this case, t = 0.800 s. Plugging in these values, we get:

d = 1/2 * 9.8 m/s^2 * (0.800 s)^2 = 3.92 meters

Therefore, the diver is 3.92 meters above the surface of the water.

(c) At what horizontal distance from the edge does the diver strike the water?

To find the horizontal distance, we need to determine how far the diver travels horizontally in 2.00 seconds (the total time for the diver to reach the water). We can use the formula:

d = v * t

In this case, v = 2.00 m/s and t = 2.00 s. Plugging in these values, we get:

d = 2.00 m/s * 2.00 s = 4.00 meters

Therefore, the diver strikes the water 4.00 meters away from the edge.

Answer 2

(a) 1.6m

(b) 6.86m

(c) 2.86s

Step-by-step explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

`x(t)=x_(0)+v_(0)t`

Where
`x(t)` is the horizontal distance at a time
`t`,
`x_(0)` is the initial horizontal position which in this case will be zero:
`x_(0)=0`. And
`v_(0)` is the initial speed:
`v_(0)=2m/s`

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:
`t=0.8s` so the horizontal distance is:

`x(0.8)=(2m/s)(0.8s)1.6m`

The answer for (a) is 1.6m

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

`y(t)=y_(0)-(1)/(2)gt^2`

Where
`y(t)` is the vertical distance at a time
`t`,
`y_(0)` is the initial vertical distance:
`y_(0)=10m` And
`g` is the gravitational acceleration:
`g=9.81m/s^2` }

(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:
`t=0.8s`

`y(0.8)=10m-(1)/(2)(9.81m/s^2)(0.8)^2`

`y(0.8)=10m-(1)/(2)(9.81m/s^2)(0.64s^2)`

`y(0.8)=10m-(1)/(2)(6.2784m)`

`y(0.8)=10m-(3.1392)`

`y(0.8)=6.86m`

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
`y(t)=y_(0)-(1)/(2)gt^2`

if
`y(t)=0`

`0=10-(1)/(2)gt^2\\-10=-(1)/(2)gt^2\\20=gt^2\\(20)/(g)=t^2\\ \sqrt{(20)/(g)}=t`

and since
`g=9.81m/s^2`

`t=\sqrt{(20)/(9.81) } =√(2.039)=1.43s`

At
`t=1.43s` the car hits the water, so the horizontal distance at that time is:

`x(t)=x_(0)+v_(0)t`

`x(1.43)=(2m/s)(1.43s)=2.86m`

the answer to (c) is 2.86s