Question

An aeroplane flying at 540 km/hr horizontally drops a packet when it was exactly above the target, at a height of 2km from the target at a height of 2km from target. by what horizontal distance, packet will miss the target.​

Answer 1

Explanation:

First convert speed 360 km/h in x m/s

360×1000/3600=100m/s

When the bomb is dropped ,it will have an initial horizontal velocity which is equal to the speed of the plane. So the bomb fall and will travel forward.

Initial horizontal velocity(v)=100m/s

Vertical velocity(u)=0m/s

Height from which bomb is dropped(h)=490m

Time=tsecs

Now from 2nd equation of motion h=ut+1/2gt²

490=0(t)+(1/2)×9.8×t²

490=4.9×t²

t²=100

t=10secs

So 10secs is the total time which bomb take to reach ground by travelling y (let) distance.

So y= vt = 100m/s × 10secs = 1000m

Distance travel by bomb =1000m or 1km