(C) Given: Height $h = 80 \ m$, initial horizontal velocity $v = 8 \ ms^{-1}$, and acceleration due to gravity $g = 10 \ ms^{-2}$.For vertical motion,

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(C) Given: Height $h = 80 \ m$, initial horizontal velocity $v = 8 \ ms^{-1}$, and acceleration due to gravity $g = 10 \ ms^{-2}$.For vertical motion, the time taken to reach the ground is given by the equation $h = \frac{1}{2}gt^2$.Substituting the values: $80 = \frac{1}{2} \times 10 \times t^2$.$80 = 5t^2 \implies t^2 = 16 \implies t = 4 \ s$.For horizontal motion, the distance $d$ covered is given by $d = v \times t$.Substituting the values: $d = 8 \times 4 = 32 \ m$.Thus, the time $t$ is $4 \ s$ and the distance $d$ is $32 \ m$.

Class 11 Physics 3-2.Motion in Plane Horizontal Projectile Motion

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From the top of a tower, $80 \ m$ high from the ground, a stone is thrown in the horizontal direction with a velocity of $8 \ ms^{-1}$. The stone reaches the ground after a time $t$ and falls at a distance of $d$ from the foot of the tower. Assuming $g=10 \ ms^{-2}$, the time $t$ and distance $d$ are given respectively by

WBJEE 2012 All WBJEE

A
$6 \ s, 64 \ m$
B
$6 \ s, 48 \ m$
C
$4 \ s, 32 \ m$
D
$4 \ s, 16 \ m$

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3-2.Motion in Plane

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Horizontal Projectile Motion

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