A ball is projected horizontally from a tower with a velocity of 4 m per second find the velocity of the ball after 0.7 seconds
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For horizontal component
v = u + at
But in horizontal acceleration = 0
So, v = u = 4m/s
Again,
Velocity in vertical direction,
v = u + at
Here , u = 0 and a = - g = -10 m/s²
So, v = -10× 0.7 = -7 m/s
In vector from final velocity = 4 i - 7j
Hence magnitude of velocity = √(4² +7²) = √65 = 8.06 m/s
v = u + at
But in horizontal acceleration = 0
So, v = u = 4m/s
Again,
Velocity in vertical direction,
v = u + at
Here , u = 0 and a = - g = -10 m/s²
So, v = -10× 0.7 = -7 m/s
In vector from final velocity = 4 i - 7j
Hence magnitude of velocity = √(4² +7²) = √65 = 8.06 m/s
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