an aeroplane flying horizontally at an altitude of for 490m with a speed of 180 kmph drops above the horizontal distance at which it hits the ground is?
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it is a formula
D=u√(2h/g)
=50√980/9.8
=50*10=500m
D=u√(2h/g)
=50√980/9.8
=50*10=500m
Answered by
1
The aeroplane is flying 490m above the ground.
The speed of aeroplane in will be the speed of block in horizontal.
The downward speed of block will be zero.
By using the formula H=ut +(1/2)gt^2 for vertical motion....
H=490m
u=0 (initial velocity in vertical direction of block)
g=9.8m/s^2
t=?
490= 0 + (9.8/2)t^2
490=4.9t^2
t^2=100
t= 10sec (because time is always +ve)
We have time taken by block to reach the ground. we also have velocity of block in horizontal direction. So we can find distance in Horizontal direction(Range).
As displacement= velocity×time
Range = 100×10
Range=1000m
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