Question

An aeroplane is flying horizontally with a constant velocity of 100kmh^(-1) at a height of 1km from the ground level.At t=0 it starts dropping packets at constant time intervals of T_(0) .If R represents the separation between two consecutive points of impact on the ground then for the first three packets R_(1)/R_(2) is
(A) 1
(B) >1
(C) <1
(D) Sufficient data is not given​

Answer 1

Answer:

(A) 1

Explanation:

Answer 2

$\frac{R1}{R2}$ for the first three packets is 1

Given : An aero plane is flying horizontally with a constant velocity of 100 km/h  

Height of the aero plane from the ground level is 1 km

Separation between two consecutive points of impact on the ground is R

To Find : $\frac{R1}{R2}$ for the first three packets

Solution :  $\frac{R1}{R2}$ for the first three packets is 1

It is given that the an aero plane is flying horizontally with a constant velocity (u)  of 100 km/h  = $\frac{500}{18}$ m/s

and , height (h)  of the aero plane from the ground level is 1 km = 1000 m

Using second equation of motion to find displacement of first impact

$s = ut + \frac{1}{2}at^{2}$

Here since velocity is constant so a is zero

s1 (displacement when the plane drops packet first time)= $\frac{500}{18}$ $t_{o}$  

s2 (displacement when the plane drops packet second time)=2× $\frac{500}{18}$ $t_{o}$

R1 separation between  s1 and s2 is s2 - s1 =    $\frac{500}{18}$$t_{o}$

s3 (displacement when the plane drops packet third time)=3× $\frac{500}{18}$ $t_{o}$

R2 separation between  s2 and s3 is s3 - s2 =    $\frac{500}{18}$$t_{o}$

$\frac{R1}{R2}$ is therefore 1 .

So  $\frac{R1}{R2}$ for the first three packets is 1

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