General Knowledge - Physics - Discussion
Discussion Forum : Physics - Section 1 (Q.No. 79)
79.
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
Discussion:
36 comments Page 4 of 4.
Sam said:
1 decade ago
Where did you get the value of "t"?
Vikash kumar pandey said:
1 decade ago
V=166.66m/s
Ht=1960m
t=20sec
R=v*t=166.66*20=3.33km.
Ht=1960m
t=20sec
R=v*t=166.66*20=3.33km.
Tms said:
1 decade ago
What is projectile motion?
Shekhar verma said:
1 decade ago
Velocity:600*5/18 m./sec.=500/3m./sec. Hy=Uy*t+1/2g*t*t : 1960=0+ 1/2*9.8*t*t therefore t=20 sec. So Range AB=v*t=500/3*20 m.=10/3 km.=3.33 km.
S RAMKIRAN said:
1 decade ago
Solution: 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.
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.
Ashu said:
1 decade ago
Could anyone explain this please?
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