A body projected vertically upwards with a certain speed from the… - Vidyadip

MCQ

A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in $t_1$. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in $t_2$. Time required to reach the ground, if it is dropped from the top of the tower, is:

✓
$\sqrt{t_1 t_2}$
B
$\sqrt{t_1-t_2}$
C
$\sqrt{\frac{t_1}{t_2}}$
D
$\sqrt{t_1+t_2}$

✓

Answer

Correct option: A.

$\sqrt{t_1 t_2}$

a
$t_1=\frac{u+\sqrt{u^2+2 g h}}{g}$

$t_2=\frac{-u+\sqrt{u^2+2 g h}}{g}$

$t=\frac{\sqrt{2 g h}}{g}$

$t_1 t_2=\frac{\left(u^2+2 g h\right)-u^2}{g^2}=\frac{2 g h}{g^2}=t^2$

$\Rightarrow t=\sqrt{t_1 t_2}$

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