Question

1. An aeroplane is flying a horizontally at a height of 400 metre with
velocity of 1000 m/sec. A packet is dropped from it so as to reach a
man standing on the ground. At what distance from the man should the
packet be dropped (take g = 10m/sec)
Ans 8953.2 metre.​

Answer 1

The required value of R = 8953.2 meter

GIVEN: An airplane is flying horizontally at a height of 400 meters with a velocity of 1000 m/sec. A packet is dropped from it to reach a man standing on the ground.
TO FIND: The distance from the man should the packet be dropped.
SOLUTION:

As we are given in the question,

An airplane is flying horizontally at a height of 400 meters with a velocity of 1000 m/sec.

Therefore,

Height(h) = 400 meters

Velocity(u) = 1000 m/sec

Also,

g = 10m/sec

As we know,

R = ut

R = u √2h/g

Therefore, putting the values in the presented formula,

R = 1000 √2×400/10

R = 1000 ×4√5

R = 4000√5

R = 4000(2.2360)

R = 8953.2

Therefore,

The required value of R = 8953.2 meter.

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Answer 2

Answer:

8953.2 meter

Explanation:

The required value of $R = 8953.2 meter$

GIVEN: plane travelling at speed of $1000 m/s$ while flying hor. height of $400 metres$. From it, a packet is released, landing near a man who is sitting on the ground.

TO FIND: the distance at which the packet should be thrown.

SOLUTION:

As specified in the query,

plane travelling at speed of $1000 m/s$ while flying hor. height of $400 metres$

Therefore,

Size ($h$) equals $400 metres$

$1000 m/s$ is the velocity ($u$).

Also,

$g = 10m/sec$

The fact is,

$R = ut$

$R$ $= u$√$2h/g$

Consequently, by entering the values in the given formula,

$R$ $= 1000$√$2*400/10$

$R$ $= 1000 *4$√$5$

$R$ $= 4000$√$5$

$R$ $= 4000(2.2360) (2.2360)$

$R$ $= 8953.2$

Therefore,

$R$ must be more than $8953.2 metres$

Velocity: Velocity is the direction at which an object is moving and serves as a measure of the rate at which its position is changing as seen from a specific point of view.

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