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 1 of 4.
Ashu said:
1 decade ago
Could anyone explain this please?
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.
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.
Tms said:
1 decade ago
What is projectile motion?
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.
Sam said:
1 decade ago
Where did you get the value of "t"?
Linda said:
1 decade ago
Actually what is this question talking about when the time is not given.
Vishnu said:
1 decade ago
v = 166.66m/s, h = 1960m.
Now s = u*t + 0.5*a*t*t which we have studied in Physics where, s = distance; u = initial velocity; a = acceleration; t = time.
In this context a = g(acceleration due to gravity).
Substituting given values in the previous equation we have,
1960 = 0+0.5*9.8*t*t(Since g=9.8m/s^2 and initial velocity,u=0).
Solving for t we get t = 20 seconds.
Now Distance = Velocity*time = 166.66(m/s)*20(seconds) = 3333.4m = 3.334km.
Now s = u*t + 0.5*a*t*t which we have studied in Physics where, s = distance; u = initial velocity; a = acceleration; t = time.
In this context a = g(acceleration due to gravity).
Substituting given values in the previous equation we have,
1960 = 0+0.5*9.8*t*t(Since g=9.8m/s^2 and initial velocity,u=0).
Solving for t we get t = 20 seconds.
Now Distance = Velocity*time = 166.66(m/s)*20(seconds) = 3333.4m = 3.334km.
Sushant Choudhary said:
1 decade ago
Distance = initial velocity *time + 1/2(acceration due to gravity * time^2).
i.e s = ut +1/2(gt^2).
For a freely falling bomb u = 0 So,
s = 1/2gt^2.
t = sqrt(2s/g) = sqrt(2*1960/9.8) = 20 seconds = 20/2600 hr.
Not the time during which bomb touches ground from the time when it was released is t and during that that time aeroplane will move a distance AB = speed of plane * time elapsed.
AB = (600)* 20/3600 = 3.33 km.
i.e s = ut +1/2(gt^2).
For a freely falling bomb u = 0 So,
s = 1/2gt^2.
t = sqrt(2s/g) = sqrt(2*1960/9.8) = 20 seconds = 20/2600 hr.
Not the time during which bomb touches ground from the time when it was released is t and during that that time aeroplane will move a distance AB = speed of plane * time elapsed.
AB = (600)* 20/3600 = 3.33 km.
Navin chahar said:
1 decade ago
What is projectile motion and not sure about this question?
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers