Question

An aeroplane is flying horizontally at a height of 980 m with velocity 100 ms 1 drops a food packet. A person on the ground is 414 m ahead horizontally from the
dropping point. At what velocity should he move so that he can catch the food packet.
1
(a) 50/2ms
50
ms
(c) 100ms 1
(d) 200 ms --

Answer 1

Explanation:

For food packet

u

h

=100 m/s

u

v

=0

s

v

=−980 m

g=−10 m/s

2

s=u

v

t+

2

1

at

2

(−980)=0+

2

1

(−10)t

2

⇒t=14 s

Distance travelled horizontally in 14 s by food packet

s

h

=u

h

×t=100×14=1400 m

Person position is 100(

2

−1) ahead of food packet at time of dropping:

=41.42 m

distance person has to run=1400−41.42 =1358.57 m in time =14 s

so his speed will be =

14

1358.57

=97 m/s

∼100 m/s

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