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.
Mustapha Daniel said:
7 years ago
Because it does not deal with height in the equation.
Matiur Rahamam said:
7 years ago
Convert velocity in horizontal direction, Ux into m/s:
Ux = 600x1000/3600 = 500/3 m/s.
Velocity of the bomb during the flight is constant due to inertia. Therefore,
Initial velocity in vertical direction, Uy = 0,
Height = 1960 m.
Step 2: Find time 't' using the equation [s = ut + 1/2at^2]
Replace distance 's' by height 'h' and acceleration 'a' by gravity 'g'
h = 1/2gt^2
t = √(2h)/g = √(2x 1960/9.8) = 20 sec.
Step 3: Find the distance traversed by the bomb in horizontal direction:
Distance AB = vx t = 500/3 x 20 = 10/3 x 10^3m = 3.333 km.
Ux = 600x1000/3600 = 500/3 m/s.
Velocity of the bomb during the flight is constant due to inertia. Therefore,
Initial velocity in vertical direction, Uy = 0,
Height = 1960 m.
Step 2: Find time 't' using the equation [s = ut + 1/2at^2]
Replace distance 's' by height 'h' and acceleration 'a' by gravity 'g'
h = 1/2gt^2
t = √(2h)/g = √(2x 1960/9.8) = 20 sec.
Step 3: Find the distance traversed by the bomb in horizontal direction:
Distance AB = vx t = 500/3 x 20 = 10/3 x 10^3m = 3.333 km.
(6)
Yoga said:
5 years ago
What is a projectile of motion?
Nas said:
3 years ago
@Tamil Nayagam,
1960/600 = 3.266.
We can take 3.3.
1960/600 = 3.266.
We can take 3.3.
(1)
Shanu said:
3 years ago
They haven't mentioned the time.
Then how did you guys calculate the projectile motion? Please explain me.
Then how did you guys calculate the projectile motion? Please explain me.
Wqwjjj said:
2 years ago
I'm not getting this. Please explain me in detail.
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