Question

From the top of a building, a particle is projected up at an angle $\theta$ with horizontal and reaches the ground in time $t_1$. A second particle projected down at an angle $\theta$ with horizontal and with same speed reaches the ground in time $t_2$. If third particle projected horizontally with same speed reaches the ground in time $t_3$, then

• A. $t_3=\sqrt{t_1{t}_2}$
• B. $t_3=\frac{t_1+t_2}{2}$
• C. $\frac{2}{t_3}=\frac{1}{t_1}+\frac{1}{t_2}$
• D. $t_3=\sqrt{t_1^2+t_2^2}$

Answer 1

The correct option is A $t_3=\sqrt{t_1{t}_2}$

$h=u\sin \theta t_1-\frac{1}{2}g{t}_1^2$.......(1)
$h=u\sin \theta t_2-\frac{1}{2}g{t}_2^2$.......(2)
$t_2\times \left(1\right)+t_1\times \left(2\right)$
$(t_1+t_2)h=-\frac{1}{2}g{t}_1{t}_2(t_1+t_2)$
$h=-\frac{1}{2}g{t}_1{t}_2$........(3)
also, $h=-\frac{1}{2}g{t}_3^2$........(4)
from (3) and (4)
$t_3^2=t_1{t}_2$
$t_3=\sqrt{t_1{t}_2}$