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).

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Solution

Here using $v = u + a t$ , where the symbols have their usual meanings,
We have:
v=V_B,
u=V_T ,
a= -9.8 m/s² ,
t= 0.5 sec,
After putting in equation we get
V_T – V_B = 4.9 m/s

Now using
$v^{2} - u^{2} = 2 a s$
Or,
V_B² – V_T²= $- 2 \times 9.8 \times 3$
= -58.8
Or we can write
(V_B-V_T)(V_B+ V_T) = -58.8 m/s
putting the value of V_T – V_B = 4.9 m/s we get
V_B + V_T = 12 m/s

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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 VT and VB, the establish a relationship between vb and vt take (g = 9.8m/s).

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