IIT JEE , AIEEE Forum - Physics , Mechanics

audittn8

An aircraft flying horizontally at 360kmph releases a bomb at a stationary tank 200m away.What must be the height of the aircraft above the tank if the bomb is to hit the tank?

dinesh_ddt

simple projectile motion problem u need to use Range=u*sqrt(2h/g)
and then fine the value of h.
in this case,
200=100*sqrt(h/5)
4=h/5
h=20m.
Am i Correct??

audittn8

The methid which I have resolved is that
1. v=0, u=100 m/s , a =g= -10 (app 10 )
Using v = u + at
t = 10 s
2. Implies,
s= ut +1/2 atsquare
s= 1000- 1/2 (10) (100)
s= 500m
For g=9.8
s= 510m
Is the method correct

dinesh_ddt

dood i think u forgot tat itz a 2D motion..!
think carefully and use the equations correctly

audittn8

But Range = u square sin 2 theta /g

dinesh_ddt

tat is oblique projectile this is a 1st order type projectile..!
eg throwing a ball from a cliff

deepamkanjani

using newton's first law of motion v=u+at
t=10 secs.
then use s=ut +1/2 a*t^2
u will get s i.e., displacement as 510 mtrs.

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audittn8

Yo ! Even I got that the Same answer. But THAT DUDE UP THERE IS CONSIDERING IT IN TERMS OF OBLIQUE PROJECTLE MOTION.EVEN I M CONFUSED

vishnu_maxlevel

dinesh is right .....the ans is 20m
it is 2d motion

karthik2007

Ans is 20m.

Now horizontal component of velocity of the bomb = 100m/s.

So time taken to cover 200m = 200/100 = 2s.

Use H = 1/2gt^2

or H = 5x4 = 20secs (assuming g = 10m/s/s)

audittn8

Thank u one and all , Sorry for the misunderstanding.

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