an aeroplane is flying horizontally with a velocity of 600kmph at a height of 1960m.when it is vertically above a point a on the ground,a bomb is released from it,the bomb strikes the point B find the distance AB
Answers
Answer:
When the bomb is dropped, it will have an initial horizontal velocity which is equal to the speed of the aeroplane. So the bomb fall and travel forward too.
initial horizontal velocity, = 600 km/h
vx= 600km/h =600*5/18 m/s =166.67 m/s
initial vertical velocity, = 0
height, h = 1960m
time taken to fall = t
h=vyt+1/2*9.8*t^2
1960=0+1/2*9.8*t^2
t=20s
In 20s, the horizontal distance travelled is AB
AB=vx*t =166.67*20 = 3333.4m
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.