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Standard IX

Physics

Derivation of Velocity-Time Relation by Graphical Method

A ball is dro...

Question

A ball is dropped from the top of a building. The ball takes $0.5$ s to fall past the length of a window, the top of the window being at a distance of $3$ m from the top of the building. If the speed of the ball at the top and the bottom of the window are $V_{T}$ and $V_{B}$ respectively, then $(g = 9.8 m / s e c^{2})$

V_{T} + V_{B} = 20 m s^{- 1}

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V_{T} - V_{B} = 4.9 m s^{- 1}

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V_{B} V_{T} = 1 m s^{- 1}

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\frac{V_{B}}{V_{T}} = 1 m s^{- 1}

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Solution

The correct option is A $V_{T} + V_{B} = 20 m s^{- 1}$
From the equations of motion we know that ${v_{t}}^{2}$ = $u^{2}$ + $2 a s$ .
here initial velocity" $u$ "=0 so $v_{t}$ = $2 \times 9.81 \times 3$ =7.67 $m / s e c$
also, $v_{b} = u + a t = 0 + 9.81$ $\times$ $0.5 = 4.905 m / s e c .$
so net velocity at the bottom of window will be
$V_{T} + V_{B}$ = $7.67 + 4.905 = 12.57 m / s e c$

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A ball is dropped from the top of a building. The ball takes 0.5sec to fall past the 3m height of a window some distance from the top of the building. If the speed of the ball at the top and the bottom is $V_{T}$ and $V_{B}$ , the establish a relationship between vb and vt take (g = 9.8m/s).

Q. A ball is dropped from the top of a building. The ball takes

0.5

s to fall past the window

3

m in length at certain distance from the top of the building. (

g = 10 m s^{- 2}

)
Speed of the ball as it crosses the top edge of the window is

Q. A ball is dropped from the top of a building. The ball takes

0.5

s to fall past the

3

m length of a window at certain distance from the top of the building. Time of travel above the window is (Take

g = 10 m / s^{2}

)

Q. A ball is dropped from the top of a building. The ball takes

0.5 s

to fall past the window of

3 m

in length at certain distance from the top of the building. (

g = 10 m / s^{2}

) How fast was the ball going as it passed the bottom of the window?

A steel ball is dropped from a building's roof and passes a window of lenght $10 m$ , taking $1 s$ to fall from the top of the window to the bottom of the window. It then falls onto a sidewalk and bounces back past the window, moving from bottom to top in $1 s$ . Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is $2 s$ . How tall is the building?

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A ball is dropped from the top of a building. The ball takes 0.5 s to fall past the length of a window, the top of the window being at a distance of 3 m from the top of the building. If the speed of the ball at the top and the bottom of the window are VT and VB respectively, then (g=9.8m/sec2)

The correct option is A VT+VB=20ms−1From the equations of motion we know that vt2=u2+2as.here initial velocity"u"=0 so vt=2×9.81×3=7.67m/secalso,vb=u+at=0+9.81×0.5=4.905m/sec.so net velocity at the bottom of window will beVT+VB=7.67+4.905=12.57m/sec