A package is dropped at time t=0 from a helicopter S that is descending steadily at a speed vi. (a) What is the speed of the package in terms of vi, g, and t? (b) What vertical distance d is it from the helicopter in terms of g and t? (c) What are the answers to parts (a) and (b) if
the helicopter is rising steadily at the same speed?
(a) When the package is dropped from the descending helicopter, it will accelerate downward due to gravity. The speed of the package can be calculated using the equation:
v = vi + gt
Where:
v is the final speed of the package
vi is the initial speed of the package (which is the same as the descending speed of the helicopter)
g is the acceleration due to gravity
t is the time elapsed
So, the speed of the package in terms of vi, g, and t is: v = vi + gt.
(b) The vertical distance d between the package and the helicopter can be determined using the kinematic equation for displacement:
d = vi*t + 0.5*g*t^2
Where:
d is the vertical distance
vi is the initial speed of the package (which is the same as the descending speed of the helicopter)
g is the acceleration due to gravity
t is the time elapsed
So, the vertical distance d in terms of g and t is: d = vi*t + 0.5*g*t^2.
(c) If the helicopter is rising steadily at the same speed, the equations for calculating the speed and vertical distance will remain the same. However, the initial speed vi will be in the opposite direction, as the helicopter is now ascending. So, the answers to parts (a) and (b) would remain the same, but with a negative initial speed (vi).
(a) To find the speed of the package in terms of vi, g, and t, we need to consider the acceleration due to gravity.
When an object is dropped, it experiences a constant acceleration of g, which is the acceleration due to gravity. In this case, as the helicopter is descending steadily, the package is also moving downward with a constant acceleration.
The equation that relates the final velocity (vf) of an object, the initial velocity (vi), the acceleration (a), and the time (t) is:
vf = vi + at
In this case, the initial velocity of the package is the same as the velocity of the descending helicopter, which is vi (given). The acceleration (a) is equal to the acceleration due to gravity, which is g. Therefore, the equation becomes:
vf = vi + gt
So, the speed of the package in terms of vi, g, and t is given by vi + gt.
(b) To find the vertical distance (d) of the package from the helicopter in terms of g and t, we can use the equation of motion:
d = vi * t + (1/2) * g * t^2
In this case, the initial vertical position (d initially) is zero because the package is dropped at time t=0. Therefore, the equation simplifies to:
d = (1/2) * g * t^2
So, the vertical distance d from the helicopter in terms of g and t is given by (1/2) * g * t^2.
(c) If the helicopter is rising steadily at the same speed, the analysis will be the same. The package will still have an acceleration due to gravity acting on it, which will be in the downward direction. However, both the initial velocity (vi) and the vertical distance (d) will have opposite signs.
For part (a), the speed of the package can still be given by the equation vf = vi + gt, where both vi and g will be negative as the package is moving upward.
For part (b), the vertical distance d from the helicopter can be given by the equation d = (1/2) * g * t^2, where g will be negative as the package is moving upward.
So, the answers to parts (a) and (b) will have the same equations, but the initial velocity and vertical distance will have opposite signs if the helicopter is rising steadily at the same speed.
no
a. V = Vi + g*t
b. d = Vi*t + 0.5g*t^2
c. V = -Vi + g*t.
d = -Vi*t + g*t