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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17
given ,
velocity of the airplane = 600 km/h = 166.6 m/sec
height from ground = 1960 meter
now ,
s = ut + 1/2 at²
1960 = 0 * t + 1/2 * 9.8 * t²
1960 = 4.9 * t²
1960 / 4.9 = t²
400 = t²
20 = t
therefore time taken = 20 sec
now the distance = velocity * time
= 166.6 * 20
=3332 meter .
therefore the answer is 3332 meter .
i hope it helps.......................^_^
velocity of the airplane = 600 km/h = 166.6 m/sec
height from ground = 1960 meter
now ,
s = ut + 1/2 at²
1960 = 0 * t + 1/2 * 9.8 * t²
1960 = 4.9 * t²
1960 / 4.9 = t²
400 = t²
20 = t
therefore time taken = 20 sec
now the distance = velocity * time
= 166.6 * 20
=3332 meter .
therefore the answer is 3332 meter .
i hope it helps.......................^_^
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0
Answer:
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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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