Physics, asked by shokumari81, 10 days ago

An aeroplane is flying horizontally at height of 490 m above a point Pon ground and a packet is to be dropped from it, which should reach at a point Q on ground which is 1600 m away from point P. What should be the speed of plane at the time of release of packet? (g=9.8 ms)​

Answers

Answered by yogeshgangwar044
13

Answer:

mean the velocity of plane must be 160 m/ sec at the time of box release.

Explanation:

given

aeroplane is flying horizontally at height of 490 m above a point P

point Q on ground which is 1600 m away from point P

when packet release from plane it start fall down with horizontal distance

first we calculate the time of falling

vertical velocity of packet is  = 0 m / sec

height fallen  h = 490 m

equation of motion

H = ut + (1/2)g t²

490 = 0+ (1/2)9.8 t²

t² = 100

t = 10 sec

means packet takes 10 sec to fall on ground

now we take horizontal motion

initial horizontal velocity u = velocity of plane = U

horizonal distance PQ = 1600 m

formula

initial horizontal velocity U = distance / time

initial horizontal velocity U =  1600/10

initial horizontal velocity U = 160 meter /sec

its mean the velocity of plane must be 160 m/ sec at the time of box release.

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