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Nezavi [6.7K]
3 years ago
10

A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the

ground, in units of m/s? Assume that air resistance is negligible. Give the answer as a positive number.
Physics
2 answers:
ddd [48]3 years ago
8 0

Answer:

32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

\frac {Vf^{2}-Vo^2}{2.a} =X

Where

Vf = final speed

Vo = Initial speed =7.3m/S

A = g=acceleration =9.81m/s^2

X = displacement =51m}

solving for Vf

Vf=\sqrt{Vo^2+2gX}\\Vf=\sqrt{7.3^2+2(9.81)(51)}=32.46m/s

the speed with the ball hits the ground is 32.46m/s

statuscvo [17]3 years ago
3 0

Answer:

The speed will be 33 m/s.

Explanation:

The equations for height and velocity of the ball are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time "t".

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

v = velocity at time "t".

Let´s place the origin of the frame of reference on the ground and consider the upward direction as positive.

First, let´s find the time it takes the ball to reach the ground:

y = y0 + v0 · t + 1/2 · g · t²

When the ball hits the ground, y = 0. Then:

0 = 51 m - 7.3 m/s · t - 1/2 · 9.8 m/s² + t²

Solving the quadratic equation:

t = 2.6 s

Now, with this time, we can calculate the velocity at the moment when the ball hits the ground:

v = v0 + g · t

v = -7.3 m/s - 9.8 m/s² · 2.6 s

v = -33 m/s

The speed will be 33 m/s.

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Lilit [14]

Answer:

The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

Explanation:

Given that,

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Angle = 19.0°

We need to calculate the component

The velocity of the ship in term x and y coordinate

v_{s_{x}}=0

v_{s_{y}}=2.0\ m/s

The velocity of the boat in term x and y coordinate

For x component,

v_{b_{x}}=v_{b}\cos\theta

Put the value into the formula

v_{b_{x}}=5.60\cos19

v_{b_{x}}=5.29\ m/s

For y component,

v_{b_{y}}=v_{b}\sin\theta

Put the value into the formula

v_{b_{y}}=5.60\sin19

v_{b_{y}}=1.82\ m/s

We need to calculate the x-component and y-component of the velocity of the cruise ship relative to the patrol boat

For x component,

v_{sb_{x}}=v_{s_{x}}-v_{b_{x}}

Put the value into the formula

v_{sb_{x}=0-5.29

v_{sb}_{x}=-5.29\ m/s

For y component,

v_{sb_{y}}=v_{s_{y}}-v_{b_{y}}

Put the value into the formula

v_{sb_{x}=2.-1.82

v_{sb}_{x}=0.18\ m/s

Hence, The x-component and y-component of the velocity of the cruise ship relative to the patrol boat is -5.29 m/s and 0.18 m/s.

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nydimaria [60]

Answer:

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Explanation:

It is given that,

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The uncertainty in momentum is 0.20%. The momentum of the bullet is given by :

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Uncertainty in momentum is,

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We need to find the uncertainty in position. It can be calculated using Heisenberg uncertainty principal as :

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\Delta x=1.07\times 10^{-33}\ m

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