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Lera25 [3.4K]
4 years ago
12

What would be the speed of an object just before hitting the ground if dropped 100 meters?

Physics
1 answer:
PolarNik [594]4 years ago
8 0

<u>We are given:</u>

the initial height of the object (h) = 100 m

initial velocity (u) = 0 m/s

we will let the value of g = 10 m/s/s

<u>Speed of the object just before hitting the ground:</u>

From the third equation of motion:

v² - u² = 2ah     (where v is the final velocity)

replacing the variables, we get:

v² - (0)² = 2(10)(100)

v² = 2000

v = 10√20 = 44.7 m/s

Therefore, the speed of the object just before hitting the ground is 44.7 m/s

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Explanation:

It is given that a particle covers 10m in first 5s and 10m in next 3s. so using the equation of motion

Case I

s=ut+

2

1

at

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10=5u+

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1

a(5)

2

20=10u+25a

4=2u+5a..............(1)

Case 2

In next 3s the particle covers more 10m distance. So

20=8u+

2

1

a(8)

2

5=2u+8a.........(2)

On solving equation (1) and (2)

4=2u+5a

5=2u+8a

a=

3

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m/s

2

Put the value of a in equation (1)

u=

6

7

m/s

Now to find distance in next 10 s. total time will be 10s

s=

6

7

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2

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×

3

1

×(10)

2

s=28.33m

Distance travelled in next 2 sec

s=28.33−20=8.33m

4 0
3 years ago
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A ball of radius R and mass m is magically put inside a thin shell of the same mass and radius 2R. The system is at rest on a ho
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Answer:

x =\frac{-R}{2}

Explanation:

From the question we are told that mass

Thin layer radius = 2R

Generally the expression for ths solution is given as

Xcm =(m*0 =m(-2R))/2m =-mR/(2m)=-R/2

the center of mass will not move at initial state  

Considering the center of mass of both bodies

xcm=\frac{m*x+m*x)}{2m} =x

x =\frac{-R}{2}

Therefore the enclosing layer moves x =\frac{-R}{2}                          

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3 years ago
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two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y
svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

v_0_x=6380 m/s

v_0_y=6770 m/s

time period of engine running=763 s

Displacement in x=4.50\times 10^6

y=7.27\times 10^6

Using s=ut+\frac{at^2}{2} in x and y direction

x=v_0_x\times t+\frac{at^2}{2}

4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}

4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}

a=-1.23 m/s^2

In y direction

y=v_0_y\times t+\frac{a't^2}{2}

7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}

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a=7.24 m/s^2

x component=-1.23 m/s^2

y component=7.24 m/s^2

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