Question

an aeroplane is flying horizontally at a height of 490m with a velocity 150ms-1. a bag containing food is to be dropped to the jawns of the ground. how far from them should the bag be dropped so that it directly reaches them?

Answer 1

so let the height R = $S_{x}$
no we know 
u = 150 m/s
$u_{y}$ = usinθ = 150sin0 = 0m/s
$u_{x}$ = ucosθ = 150cos0 = 150m/s
s = ut + 1/2at²
so 
$S_{y} = u_{y}t + 1/2a_{y}t^{2}$
⇒490 = 0 + 1/2 x 10 x t²
⇒ t = √98 sec = 9.9 sec ≈ 10 sec
again
$S_{x} = u_{x}t + 1/2g_{x}t^{2}$
R = 150 x t + 0 (along x axis gravoty is 0m/s)
R = 150 x 10
R = 1500 m
so it must be at a dropped 1500 m before the point

Answer 2

for vertical component of motion:
s=1/2gt²
t=√(2s/g) =√(2*490)/9.8=√100=10 second (time taken to reach the ground)
horizontal component of motion:
distance travelled in 10 seconds =10* 150
                                                =1500 meters
The bag should be dropped when the plane is 1500 away from the target.