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`        A bal is thrown horizontally at 9.8m/sec from a tower 40m high.The resultant velocity after 1 sec is....m/sec.`
7 years ago

105 Points
```										Dear chakrala,
In a horizontal projectile the horizontal velocity is constant and the vertical velocity changes because of the accelaration due to gravity
Hence Vx = 9.8 m/s
the verticla veloctiy after 1 sec is
Vy = -g*1      (the intial velocity in the verticla direction is zero)
Vy = -9.8 m/s
Hence the resultant velocity after 1 sec is
V = 9.8 m/s
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```
7 years ago
510 Points
```										velocity in horizontal direction remains same
so Vx=9.8m/s which is constant
initially velocity in vertical direction is zero
velocity after 1 sec is Vy=Uy+at=0+9.8t            (a=g)
Vy=9.8m/s
V=9.8i-9.8j in vector form
=9.8sqrt2 in magnitude
```
7 years ago
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