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12 Jul 2009 23:42:03 IST

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a ball is dropped from a height. if it takes 0.200s to cross the last 6.00m before hitting the g

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a ball is dropped from a height. if it takes 0.200s to cross the last 6.00m before hitting the ground, find the height from which it was dropped.

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chandramouli bandhopadhya

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13 Jul 2009 00:51:05 IST

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the answer is 32.54658465468454 metres...

BALGANESH

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13 Jul 2009 09:39:49 IST

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let the intial velocityof th eball b4 the last 6m be =U

a=10 s=6 t=0.2

s=ut+1/2at^2

6=0.2 U+1/2*10*0.2*0.2

0.2U=5.8 U=29

now consider the journey from rest to this height

V=29m/s U=0 a=10

v^2=u^2+2as

29*29=2*10*s

therefore s= 29*29/20 = 42.05

thereforethe total height = 42.05+6 = 48.05m

Siddharth Kothari

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13 Jul 2009 09:52:08 IST

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You don't the vel. the ball just 0.200s before hitting the ground.

Let it be U. (Taking downward dir. as -ve and upward as +ve) :sign convention

d = -6m

t = 0.2s

a = -10ms^-2

Substituting in the equation for constant accelⁿ. : d = ut + 1/2*a*t²

-6 = u(0.2) + 1/2(-10)(o.2)²

-60 = 2u -2

2u = -58

u = -29ms^-2

This is the velocity 0.2 seconds before the fall.

Siddharth Kothari

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13 Jul 2009 09:56:06 IST

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final vel . v = -29 + (-10)(0.2)

v = -31m/s

Initial velocity is 0 as the ball is dropped.

So, v² - u²= 2as

(-31)2 = 2(-10)s

961 = -20s

s = -(961)/20

s = -48.05m

This is the displacement, and negative sign shows that it's direction is downwards.

Hence, height of the tower is 48.05m

Siddharth Kothari

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13 Jul 2009 09:57:57 IST

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Balganesh, your answer was right.

But, as displacement, velocity and acclⁿ. are vector quantities, a direction is necessary.

BALGANESH

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13 Jul 2009 09:59:26 IST

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u r right

but here its about height

i dont think its necessary to add teh signs

it just confuses ppl

so didnt use them

anyways urs is absolutely correct

edison

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13 Jul 2009 12:10:45 IST

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Let height from which ball is dropped = H

If we consider intial velocity of ball during the last 6m distance travelled by ball to be = u

Then using,

s=ut+1/2at²

where a = g (acceleration due to gravity) =10 m/s²

t = 0.2 s

Thus 6 = 0.2 u + 0.5 * 10 * 0.04

or u = 29m/s

Now to acquire the velocity of 29m/s the ball should be dropped from height 'x' under the influence of 'g' and

thus we can make use the following equation

v² - u² = 2gs

or v² - u² = 2gx

or 29² = 2*10*x

or x = 42.05 m

Thus total height from which the ball was dropped is given by

H = x + 6

or H = 42.05 + 6 = 48.05 m

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