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Standard IX

Physics

Derivation of Position-Time Relation by Graphical Method

A marble with...

Question

A marble with speed 20 cm/s rolls off the edge of a table 80 cm high. How long does it take to drop to the floor? How far horizontally from the table edge does the marble strike the floor?

8.4 m

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8.0 cm

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8.0 m

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8.9 cm

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Solution

The correct option is B
8.0 cm

$v_{o x} = v_{0} = 20 c m / s$

$v_{o y = 0}$

$a_{y} = + g = + 980 c m / s^{2}$

$a_{x} = 0$

To find time of fall, $y = v_{0 y t} + \frac{1}{2} g t^{2}$

$80 c m = 0 + (490 c m / s^{2}) t^{2}$

$t = 0.40 s e c$

$x = v_{o x} t = (20 c m / s) (0.40 s) = 8.0 c m$

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A marble with speed 20 cm/s rolls off the edge of a table 80 cm high. How long does it take to drop to the floor? How far horizontally from the table edge does the marble strike the floor?

The correct option is B 8.0 cm vox=v0=20 cm/s voy=0 ay=+g=+980 cm/s2 ax=0 To find time of fall, y=v0yt+12gt2 80 cm=0+(490 cm/s2)t2 t=0.40 sec x=voxt=(20 cm/s)(0.40 s)=8.0 cm

The correct option is B
8.0 cm

$v_{o x} = v_{0} = 20 c m / s$

$v_{o y = 0}$

$a_{y} = + g = + 980 c m / s^{2}$

$a_{x} = 0$

To find time of fall, $y = v_{0 y t} + \frac{1}{2} g t^{2}$

$80 c m = 0 + (490 c m / s^{2}) t^{2}$

$t = 0.40 s e c$

$x = v_{o x} t = (20 c m / s) (0.40 s) = 8.0 c m$