(D) The initial velocity of the aeroplane is horizontal,so the initial vertical component of the velocity of the packet is $u_y = 0 \ m/s$.Using the s

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(D) The initial velocity of the aeroplane is horizontal,so the initial vertical component of the velocity of the packet is $u_y = 0 \ m/s$.Using the second equation of motion in the vertical direction: $s_y = u_y t + \frac{1}{2} a_y t^2$.Here,$s_y = h$,$u_y = 0$,and $a_y = g$.Substituting these values: $h = 0 \cdot t + \frac{1}{2} g t^2$.$h = \frac{1}{2} g t^2$.Solving for $t$: $t^2 = \frac{2h}{g}$.Therefore,$t = \sqrt{\frac{2h}{g}}$.

Class 11 Physics 3-2.Motion in Plane Projectile Motion from Hight

Easy

An aeroplane is moving with a velocity $u$. It drops a packet from a height $h$. The time $t$ taken by the packet in reaching the ground will be

A
$\sqrt{\frac{2g}{h}}$
B
$\sqrt{\frac{2u}{g}}$
C
$\sqrt{\frac{h}{2g}}$
D
$\sqrt{\frac{2h}{g}}$

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3-2.Motion in Plane

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Projectile Motion from Hight

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$A$ ball is projected from a point $O$ as shown in the figure. It will strike the ground after ........ $s$ $\left(g=10 \, m/s^2\right)$.

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An aeroplane is moving with a velocity $u$. It drops a packet from a height $h$. The time $t$ taken by the packet in reaching the ground will be

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