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2.Motion in Straight Line
hard
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:
A$\sqrt{t_1 t_2}$
B$\sqrt{t_1-t_2}$
C$\sqrt{\frac{t_1}{t_2}}$
D $\sqrt{t_1+t_2}$
(JEE MAIN-2024)
Solution
$t_1=\frac{u+\sqrt{u^2+2 g h}}{g}$
$t_2=\frac{-u+\sqrt{u^2+2 g h}}{g}$
$t=\frac{\sqrt{2 g h}}{g}$
$t_1 t_2=\frac{\left(u^2+2 g h\right)-u^2}{g^2}=\frac{2 g h}{g^2}=t^2$
$\Rightarrow t=\sqrt{t_1 t_2}$
$t_2=\frac{-u+\sqrt{u^2+2 g h}}{g}$
$t=\frac{\sqrt{2 g h}}{g}$
$t_1 t_2=\frac{\left(u^2+2 g h\right)-u^2}{g^2}=\frac{2 g h}{g^2}=t^2$
$\Rightarrow t=\sqrt{t_1 t_2}$
Standard 11
Physics
