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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

VladimirAG [237]

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:

$x(t)=x_{0}+v_{0}t$

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:

$y(t)=y_{0}-\frac{1}{2}gt^2$

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: $t=0.8s$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.8)^2$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.64s^2)$

$y(0.8)=10m-\frac{1}{2}(6.2784m)$

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

$y(0.8)=6.86m$

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 $y(t)=y_{0}-\frac{1}{2}gt^2$

if $y(t)=0$

$0=10-\frac{1}{2}gt^2\\-10=-\frac{1}{2}gt^2\\20=gt^2\\\frac{20}{g}=t^2\\ \sqrt{\frac{20}{g}}=t$

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

$t=\sqrt{\frac{20}{9.81} } =\sqrt{2.039}=1.43s$

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

6 0

3 years ago

Controlled variables

PilotLPTM [1.2K]

Answer:

A control variable in scientific experimentation is an experimental element which is constant and unchanged throughout the course of the investigation.

Explanation:

7 0

3 years ago

2 Jim keeps a can on the ramp. The can

IceJOKER [234]

Answer:

The description of the statement is summarized throughout the explanation section below.

Explanation:

The layer of the ramp must've been harsh, which exacerbated the toy movement to occur gradually. This same friction force was indeed starting to function on the remote control car as well as trying to stop that one to react quickly.
Jim does have to simplify the exterior including its ramp by scrubbing or greasing to overcome impact as well as make driving travel quicker.

6 0

3 years ago

If a car was moving at a speed of 25mph for 6 hours, how far would the car travel?

Anestetic [448]

Answer:

150 miles

Explanation:

25mph x 6hrs = 150miles

7 0

3 years ago

A 87 kg man has a total mechanical energy of 1780 J .If he is swinging downward and is currently 1.4 m above the ground, what is

borishaifa [10]

Answer:

6.4m/s

Explanation:

The total mechanical energy of the man is 1780J.

This mechanical energy is the energy due to the motion of the body and it is a form of kinetic energy.

Also, mass = 87kg

Kinetic energy = $\frac{1}{2}$ m v²

m is the mass

v is the velocity

1780 = $\frac{1}{2}$ x 87 x v²

v² = 40.9

v = 6.4m/s

4 0

3 years ago