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Physics-

General

Easy

Question

A ball is dropped from a high rise platform at $t equals 0$ starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed $v$ . The two balls meet at $t equals 18 s$ . What is the value of $v$ ? (taking $g equals 10 blank m divided by s to the power of 2 end exponent$ )

$60 m / s$
$75 m / s$
$55 m / s$
$40 m / s$

The correct answer is: $75 blank m divided by s$

Let two balls meet at depth $h$ from platform
So $h equals fraction numerator 1 over denominator 2 end fraction g open parentheses 18 close parentheses to the power of 2 end exponent equals v open parentheses 12 close parentheses plus fraction numerator 1 over denominator 2 end fraction g open parentheses 12 close parentheses to the power of 2 end exponent rightwards double arrow v equals 75 blank m s to the power of negative 1 end exponent$

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A balloon gong upward with a velocity of $12 blank m s to the power of negative 1 end exponent$ is at a height of 65 m from the earth’s surface at any instant. Exactly at this instant a ball drops from it. How much time will the ball take in reaching the surface of earth?
$open parentheses g equals 10 blank m s to the power of negative 2 end exponent close parentheses$

A balloon gong upward with a velocity of $12 blank m s to the power of negative 1 end exponent$ is at a height of 65 m from the earth’s surface at any instant. Exactly at this instant a ball drops from it. How much time will the ball take in reaching the surface of earth?
$open parentheses g equals 10 blank m s to the power of negative 2 end exponent close parentheses$

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The graph of displacement-time for a body travelling in a straight line is given. We can conclude that

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A particle moves along a straight line such that its displacement at any time $t$ is given by $S equals t to the power of 3 end exponent minus 6 t to the power of 2 end exponent plus 3 t plus 4 blank m e t r e s$
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A body is moving along a straight line path with constant velocity.At an instant of time the distance travelled by it is S and its displacement is D, then

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With what velocity a ball be projected vertically so that the distance covered by it in 5^th second is twice the distance it covers in its 6^th second ( $g equals 10 blank m divided by s to the power of 2 end exponent$ )

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Physics-General

A bullet fired into a fixed target loses half of its velocity after penetrating $3 blank c m$ . How much further it will penetrate before coming to rest assuming that if faces constant resistance to motion

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The displacement-time graphs of two particles $A$ and $B$ are straight lines making angles of respectively $30 degree$ and $60 degree$ with the time axis. If the velocity of $A$ is $v subscript A end subscript$ and that of $B$ is $v subscript B end subscript$ , then the value of $fraction numerator v subscript A end subscript over denominator v subscript B end subscript end fraction$ is

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Which of the following is true

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A particle moving with a uniform acceleration travels 24 $m$ and 64 $m blank$ in the first two consecutive intervals of 4 $s e c$ each.
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Physics-General

A balloon is rising vertically up with a velocity of $29 blank m s to the power of negative 1 end exponent$ . A stone is dropped from it and it reaches the ground in 10 seconds. The height of the balloon when the stone was dropped from it is $left parenthesis g equals 9.8 blank m s to the power of negative 2 end exponent right parenthesis$

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Equation of displacement for any particle is $blank s equals 3 t to the power of 3 end exponent plus 7 t to the power of 2 end exponent plus 14 t plus 8 m$ . Its acceleration at time $t equals 1 blank s e c$ is

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