Complete Question
In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
Answer:
The speed of the helicopter is [tex]u = 7.73 \ m/s [/tex]
Explanation:
From the question we are told that
The height at which he let go of the brief case is h = 130 m
The time taken before the the brief case hits the water is t = 6 s
Generally the initial speed of the briefcase (Which also the speed of the helicopter )before the man let go of it is mathematically evaluated using kinematic equation as
[tex]s = h+ u t + 0.5 gt^2[/tex]
Here s is the distance covered by the bag at sea level which is zero
[tex]0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2[/tex]
=> [tex]0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2[/tex]
=> [tex]u = \frac{-130 + (0.5 * 9.8 * 6^2) }{6}[/tex]
=> [tex]u = 7.73 \ m/s [/tex]
