An aeroplane is flying horizontally with a velocity of 600 km/h and at a height of 1960 m. When it is vertically at a point A on the ground a bomb is released from it. The bomb strikes the ground at point B. The distance AB is
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Explanation:
The velocity with which the bomb is released is 600km/h(horizontal).
Vertical velocity of the bomb is zero.
So.distance AB=600×18/5×t
t is the time taken by the bomb to reach the ground.
1960=g×2t/2
which gives AB equal to 3.3.km.
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Explanation:
see in the question its given:
Velocity= 600Km/hr = 600x1000/3600 m/s=166.66m/s
Ht=1960m.
To Find: The range of the projectile (R).
Its actually a Projetile motion .
For a projectile Ht=1/2*g*t*t.There fore we get t= 20s.
R=V*t=166.66*20=3333.33m/s=3.333Km/hr.
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