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Standard IX

Physics

Second Equation of Motion

An object thr...

Question

An object thrown upwards with a velocity $u$ from the top of a tower of height $H$ reaches the ground in $t_{1}$ seconds. When thrown downwards with the same speed it reaches the ground in $t_{2}$ seconds. When dropped from the top of tower it reaches in $t_{3}$ seconds. Then $t_{3}$ is equal to (in terms of $t_{1}$ and $t_{2}$ )

t_{1}

t_{2}

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\sqrt{t_{1} t_{2}}

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t_{1}

t_{2}

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t_{1}

t_{2}

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Solution

The correct option is B $\sqrt{t_{1} t_{2}}$
assuming downward direction as positive we get
$H = - u t_{1} + \frac{1}{2} a t_{1}^{2}$
and $H = u t_{2} + \frac{1}{2} a t_{2}^{2}$
multiplying 1 with $t_{2}$ and 2 with $t_{1}$ and then add both we get
$=> H (t_{1} + t_{2}) = \frac{1}{2} a t_{1} t_{2} (t_{1} + t_{2})$
$=> 2 H = a T_{1} . t_{2}$
so when ball is simply released from top we get
$H = \frac{1}{2} a t_{3}^{2}$
using value of H we get
$=> t_{3}^{2} = t_{1} . t_{2}$
so best answer is option B.

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An object thrown upwards with a velocity u from the top of a tower of height H reaches the ground in t1 seconds. When thrown downwards with the same speed it reaches the ground in t2 seconds. When dropped from the top of tower it reaches in t3 seconds. Then t3 is equal to (in terms of t1 and t2)