Question

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)​

Answer 1

Answer: 3.33km

Explanation: From h=12gt2

we have, tOB=2hOAg−−−−−√=2×19609.8−−−−−−−−√=20s

Horizontal distance AB=vtOB

=(600×518m/s)(20s)

=3333.33m=3.33km.

Answer 2

Given:

An airplane is flying horizontally at a height of 490m. above a point P on the ground.

To Find:

Speed of plane at the time of the release of a packet?

Explanation:

• The height of the airplane from point P on the ground is 490m

        and we have to find the speed of the plane.

• By the third equation of motion.

        $s=ut+\frac{1}{2} at^2$

       Here u is the initial velocity of the plane which is 0.

       $490=\frac{1}{2} \times 9.8 \times t^2$

        $t^2=\frac{2\times 490}{9.8}$

        $t=10s$

• Distance between the point P on the ground and point Q is 1600m. it means the distance traveled by airplane in 10 seconds is 1600 meters.

• By the formula of speed, speed can be calculated by distance traveled per unit time.

        $Speed=\frac{Distance }{time}$

        $Speed= \frac{1600}{10}=160 m/s$

So, the speed of an airplane is 160m/s.