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Gennadij [26K]
2 years ago
7

During several practice runs you stand at the end of a diving board and launch the basketball horizontally from a position 4.50

m above the water. If the average landing spot is 6.25 m horizontally from your initial position, what is the average launch speed?

Physics
1 answer:
mr Goodwill [35]2 years ago
3 0

Answer:

6.52m/s

Explanation:

Hello!

To solve this problem follow the steps below, the complete procedure is in the attached image

1. draw a complete outline of the problem involving forces and geometries

2. Find the time that elapses from when the ball is thrown until it hits the water, this is achieved using the kinematic equation for constant acceleration and zero vertical initial velocity.

3. taking into account that the ball is launched horizontally, and that there is no horizontal acceleration, the only thing we must do is find the speed by dividing the distance traveled horizontally and dividing it by time

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Driving your Ferrari through the Italian countryside at a speedy 88 m/s, you approach an opera diva singing a high C (1,046 Hz).
MrRissso [65]

Answer:

You will hear the note E₆

Explanation:

We know that:

Your speed = 88m/s

Original frequency = 1,046 Hz

Sound speed = 340 m/s

The Doppler effect says that:

f' = \frac{v \pm v0 }{v \mp vs}*f

Where:

f = original frequency

f' = new frequency

v = velocity of the sound wave

v0 = your velocity

vs = velocity of the source, in this case, the source is the diva, we assume that she does not move, so vs = 0.

Replacing the values that we know in the equation we have:

f' = \frac{340 m/s + 88m/s}{340 m/s} *1,046 Hz = 1,316.73 Hz

This frequency is close to the note E₆ (1,318.5 Hz)

7 0
2 years ago
A cannon fires a shell straight upward; 2.3 s after it is launched, the shell is moving upward with a speed of 17 m/s. Assuming
PtichkaEL [24]

Answer:

The speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.                    

Explanation:

Let u is the initial speed of the launch. Using first equation of motion as :

u=v-at

a=-g

u=v+gt\\\\u=17+9.8\times 2.3\\\\u=39.54\ m/s

The velocity of the shell at launch and 5.4 s after the launch is given by :

v=u-gt\\\\v=39.54-9.8\times 5.4\\\\v=-13.38\ m/s

So, the speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.

6 0
3 years ago
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