Question

an aeroplane moving horizontally with a speed of 180 km/hr drops a packet while flying at a height of 490m. The horizontal range is

Answer 1

Since the plane is moving with a horizontal velocity of 180km/hr, the packet also aquires that horizontal velocity. Hence, when the packet is being dropped, its horizontal velocity is 180km/hr and its vertical velocity is 0 km/hr.

Now, let us analyse the horizontal direction.
We can apply the second equation of motion in this direction.
Here, the horizontal acceleration of the packet is 0m/s².
R = ut +at²/2
Since, a =0 and u = 180km/hr (remember its the horizontal velocity)
R = 180×t
If we somehow know the value of this t, we can easily calculate the horizontal range. This time taken by the packet to cover whole horizontal range is also the same as the time taken by it to cover the vertical range or simply height.

So, we now just need to analyse the vertical direction now and calculate this time t.

For calculating this too, we can use the 2nd equation of motion, ie

s = ut +at²/2
here s = H = vertical height
a = g = 9.8m/s²
u = 0m/s= initial vertical velocity of packet is zero

So, H = gt²/2
t² = 2H/g = 2 × 490/9.8 = 100
t = 10 sec

Now, we have got the time and we just need to substitute this in the range formula, ofcourse after fixing the units.

R = 180 km/hr ×t
= 180 ×5/18 × t
= 50m/s × t
= 50 m/s × 10s = 500m = 0.5 km

I hope you understand the concepts.
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