Answers

2014-08-15T15:26:00+05:30
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
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2014-08-15T17:28:30+05:30
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.
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