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
1200 m
0.33 km
3.33 km
33 km
Answer: Option
Explanation:
No answer description is available. Let's discuss.
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.
(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.
(1)

Shanu said:   3 years ago
They haven't mentioned the time.

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.


Post your comments here:

Your comments will be displayed after verification.