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 3 of 4.
Rajat shukla said:
10 years ago
Why is initial velocity 0? Why not 600 as plane is moving at 600?
Tamil nayagam.g said:
1 decade ago
1960/600 = 33.2666666666667.
Tushar said:
1 decade ago
When a body is projected into space and no longer propelled by fuel and other external forces except gravity it's motion is called a projectile motion.
Umar said:
1 decade ago
SOLUTION:
Velocity = 600km/hr=600*1000/3600m/s.
= 166.66m/s.
FORMULA:
Ht = 0.5*a*t*t.
Ht = 0.5*9.8*2*2.
Ht = 19.6+0.4.
H t= 20 seconds.
R = V*t.
R = 166.66*20 SEC.
R = 333.2km.
R = 333.2/1000.
R = 3.33km.
Therefore R is 3.33km.
Velocity = 600km/hr=600*1000/3600m/s.
= 166.66m/s.
FORMULA:
Ht = 0.5*a*t*t.
Ht = 0.5*9.8*2*2.
Ht = 19.6+0.4.
H t= 20 seconds.
R = V*t.
R = 166.66*20 SEC.
R = 333.2km.
R = 333.2/1000.
R = 3.33km.
Therefore R is 3.33km.
Vidya sagar maddheshiya said:
1 decade ago
BOMB WILL FALL UNDER THE EFFECT OF EARTH GRAVITY(9.8M/SEC SQUIRE) AND AEROPLANE WILL MOVE IN HORIZONTAL DIRECTION WITH ITS HORIZONTAL SPEED V(V=u+a*t).
Here the value of V will be 1600 KmpH(=166.66m/s) because its initial speed(u) is zero. The taken by bomb to reach on earth will be calculated by formula (S=ut+0.5at*t); here the S is the height and a is equal to earth gravity(g=9.8) and u=0, so formula will be H=0.5gt*t and thus value of t( time taken by bomb to reach on earth will be) will be 20 sec.
Now the bomb will move in the horizontal direction in 20 sec and with the speed( same speed of aeroplane) of 1600(=166.66m/s) and it will move in horizontal direction (distance = speed x time) by 166.66x20 sec.
Here the value of V will be 1600 KmpH(=166.66m/s) because its initial speed(u) is zero. The taken by bomb to reach on earth will be calculated by formula (S=ut+0.5at*t); here the S is the height and a is equal to earth gravity(g=9.8) and u=0, so formula will be H=0.5gt*t and thus value of t( time taken by bomb to reach on earth will be) will be 20 sec.
Now the bomb will move in the horizontal direction in 20 sec and with the speed( same speed of aeroplane) of 1600(=166.66m/s) and it will move in horizontal direction (distance = speed x time) by 166.66x20 sec.
Yaima said:
1 decade ago
How come the time is 20 Seconds?
Navin chahar said:
1 decade ago
What is projectile motion and not sure about this question?
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.
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.
Linda said:
1 decade ago
Actually what is this question talking about when the time is not given.
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