Question

A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the following. (a) What is the speed of the package in terms of v0, g, and t? (Use any variable or symbol stated above as necessary. Let down be positive.) (b) What distance d is it from the helicopter in terms of g and t? d = (c) What are the answers in parts (a) and (b) if the helicopter is rising steadily at the same speed? speed distance d =

Answer 1

Answer:

Part a)

$v = \sqrt{v_o^2 + g^2t^2}$

Part b)

$d = \frac{1}{2}gt^2$

Part c)

$v_f = v_o - gt$

Part d)

$d = \frac{1}{2}gt^2$

Explanation:

Part a)

As we know that speed of package is same as that of helicopter in horizontal direction

So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall

So we have

$v_y = -gt$

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{v_o^2 + g^2t^2}$

Part b)

Distance from helicopter is same as the distance of free fall

so we will have

$d = \frac{1}{2}gt^2$

Part c)

If helicopter is rising upwards with uniform speed

then final speed of the package after time t is given as

$v_f = v_i + at$

$v_f = v_o - gt$

Part d)

distance from helicopter

$d = \frac{1}{2}gt^2$

Answer 2

Answer:

a)$v = v_{0}+ gt$

b)$s = v_{0} + \frac{1}{2} gt$

c) $v = v_{0} -gt\\s= v_{0}t -\frac{1}{2} gt$

Explanation:

a) final speed  of the package:

Initial speed = v₀

Time taken by the package = t

let final speed be given by v.

This is given by:

$v = u + at\\ = v_{0} + gt$

b)The distance is given as:

$S = ut + \frac{1}{2} at^{2}$

which is equals to:

$s = v_{0} t+\frac{1}{2} gt$

c) if the helicopter is rising, the acceleration g will be negative, so the equations will become:

$v= v_{0} -gt$ and $s=v_{0} t - \frac{1}{2} gt$