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A ball is dropped from top of the building.The ball takes 0.5s to fall past 3m of a window some distance fro the top of the building. If the speedof the ball at top and bottom of the vindow are’ U’ and’V’ respectively then ‘U-V’=?, V+V=? UV=?
A ball is dropped from top of the building.The ball takes 0.5s to fall past 3m of a window some distance fro the top of the building. If the speedof the ball at top and bottom of the vindow are’ U’ and’V’ respectively then ‘U-V’=?, V+V=? UV=?

```
4 years ago

```							V2-U2=2g(3) also, V=U+g(0.5) i.e V-U=0.5g(V+U)(V-U)=6g(V+U)(0.5g)=6gV+U=12,U-V=-4.9, Q. Must be modified as,A ball is dropped from top of the building.The ball takes 0.5s to fall past 3m of a window some distance fro the top of the building. If the speedof the ball at top and bottom of the vindow are’ U’ and’V’ respectively then ‘U-V’=?, V+U=? UV=?
```
4 years ago
```							Let initial velocity at top of window be 'u' and velocity at bottom be ' v' Use S = ut + 1/2at2   where u is initial velocity a =g(10) t =0.5  S =3 ,. We get value of u as 3.5m/S. Now use v = u + at  where u is initial velocity and a =g,t =0.5 you will get the value of v now use it to find u-v ,v+v ,u.v
```
4 years ago
```							Here V, S, and U is given... So,  use               V=U+GT     ;         (U+V) t=2S              V-U=GT       ;NOW PUT THE VALUES YOU WILL GET PERFECT ANSWERS.... U+V=12 AND U-V=-4.9
```
one year ago Yash Chourasiya
256 Points
```							Hello StudentLet initial velocity at top of window be 'U' and velocity at bottom be ' V'V2 – U2 = 2aSV2- U2= 2g(3) ….........(1)also,V = U + atV = U + g(0.5)V – U = 0.5g........(2)Equation (1) can also be written as(V+U)(V-U)=6gPutting value of V – U from eq. (2), we get(V+U)(0.5g)=6gV + U = 12 m/sec …...(3)In equation (2) putting value of g, and we getU – V = -4.9 m/sec …....(4)From equation (3) & (4), we getUV = 29.9975 = 30 aproxI hope this answer will help you.
```
7 months ago
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