an aeroplane movug horizontal with speed 720 km/h drops a food packet,while flying at height of 396.9m the time taken by the food packet to reacg the ground and its horizontal range is what?
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The distance S=396.9 metres.
time t=?
g= 9.8 m/sec²
Initial velocity u =0
Using the first equation of motion :-
S=ut+1/2gt²
⇒396.9=0+1/2×9.8t²
⇒t²=396.9/4.9
⇒t=√81
⇒t=9 seconds. Ans.
The velocity 720 km/h will not be used as it is the velocity of the aeroplane and of the food packet till it is inside the aeroplane but when it is removed from the aeroplane the food packet comes to rest immediately by the property of inertia and then it begins to fall down with an acceleration g.
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time t=?
g= 9.8 m/sec²
Initial velocity u =0
Using the first equation of motion :-
S=ut+1/2gt²
⇒396.9=0+1/2×9.8t²
⇒t²=396.9/4.9
⇒t=√81
⇒t=9 seconds. Ans.
The velocity 720 km/h will not be used as it is the velocity of the aeroplane and of the food packet till it is inside the aeroplane but when it is removed from the aeroplane the food packet comes to rest immediately by the property of inertia and then it begins to fall down with an acceleration g.
I hope u got it!
Plz mark as the best if u liked it!
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