Respuesta :
Answer:
The horizontal distance the package traveled while falling, S = 197.99 m
Explanation:
Given,
The horizontal velocity of the helicopter, v = 70 m/s
The time taken by the package to fall into the water is, t = 8.0 s
The initial horizontal component of velocity, Vx = 70 m/s
The initial vertical component of velocity, Vy = 0 m/s
The time of flight is given by the formula
[tex]t=\frac{Vy+\sqrt{Vy^{2} +2gh}}{g}[/tex]
Since Vy = 0 ; solving for h
h = gt²/2 meter
Substituting the given values in the above equation
h = (9.8 x 8²) / 2
= 39.2 m
The horizontal distance traveled while falling is given by the formula
[tex]S=\frac{Vx[Vy+\sqrt{Vy^{2} +2gh}] }{g}[/tex]
Since, Vy = 0, equation becomes
[tex]S=\frac{Vx\sqrt{2gh} }{g}[/tex]
Substituting the values in the above equation
[tex]S=\frac{70\sqrt{2X9.8X39.2} }{9.8}[/tex]
S = 197.99 m
Hence, the horizontal distance the package travels while falling is, S = 197.99 m
Answer: D = 560m
Explanation: The helicopter is traveling with a velocity of 70m/s in the x-axis, and we know that it takes 8 seconds to fall into the water.
Whem the helicopter "throws" the package, the only equations for the package are the initial velocity, that is the same of the helicopter (70 m/s in the x axis)
and the gravity acceleration in the y axis, but we only care for the horizontal distance, so this does not matter.
Then, knowing that the package travesl for 8 seconds with a velocity of 70m/s in the horizontal direction, the total distance traveled is:
70m/s*8s = 560m
