From certain height, two bodies are projected horizontally with velocities $10 m / s$ and $20 m / s$ . They hit the ground in $t_{1}$ and $t_{2}$ seconds. Then

t_{1} = t_{2}

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t_{1} = 2 t_{2}

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t_{2} = 2 t_{1}

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t_{1} = \sqrt{2 t_{2}}

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Solution

The correct option is A $t_{1} = t_{2}$
let $h$ be the height from which two bodies are projected
using 2nd equation of motion in y direction:
$h = u_{y} t + \frac{g t^{2}}{2}$
$u_{y} = 0$ for both case
$t = \sqrt{\frac{2 h}{g}}$
since $t$ is independent of $u_{x}$
so $t_{1} = t_{2}$

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From certain height, two bodies are projected horizontally with velocities 10m/s and 20m/s. They hit the ground in t1 and t2 seconds. Then

The correct option is A t1=t2let h be the height from which two bodies are projected using 2nd equation of motion in y direction:h=uyt+gt22uy=0 for both caset=√2hgsince t is independent of uxso t1=t2

From certain height, two bodies are projected horizontally with velocities $10 m / s$ and $20 m / s$ . They hit the ground in $t_{1}$ and $t_{2}$ seconds. Then

The correct option is A $t_{1} = t_{2}$
let $h$ be the height from which two bodies are projected
using 2nd equation of motion in y direction:
$h = u_{y} t + \frac{g t^{2}}{2}$
$u_{y} = 0$ for both case
$t = \sqrt{\frac{2 h}{g}}$
since $t$ is independent of $u_{x}$
so $t_{1} = t_{2}$