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
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.