Can anyone tell me what equation to use to solve this problem? I am having trouble. Thanks?

Question

A helicopter is ascending vertically with a speed of 5.20m/s. At a height of 125m above the earth, a package is dropped from a window. How much time does it take for the package to reach the ground? [ hint: the package's intial spped equals the helicopter's.]

Answer 1

same as your last question, this is a motion problem and there are four possible equations you need to know for these types of problems

(1) Vf = Vi + at
(2) d = 1/2 (Vf + Vi) x t
(3) d = Vit + 1/2 a t^2
(4) Vf^2 = Vi^2 + 2ad

where Vf = velocity final, Vi= velocity initial, a = acceleration, t = time, d = distance

for this problem you're given Vi = 5.20 m/s, d = 125 m and you know a = 9.8 m/s^2. You need to find "t"

equations (1) and (2) above have a Vf term which you don't know right?

from equation (3).....

d = Vit + 1/2 a t^2
125 m = 5.20 m x t + 1/2 x 9.8 m/s^2 x t^2

4.9t^2 + 5.20 t -125 = 0

solve for t....

hint use quadratic formula. I get t = 4.55 seconds


*******update*******

after reading the joy's answer, I realized I made and err calculating the answer. the answer should be 4.55 seconds. not 5.61

Answer 2

Start timing (let t=0) when you release the package.
When t=0, the height of the package is 125 meters, and
the copter is carrying it upward at 5.2 meters per second
so its height is 125 + 5.2t, minus half the acceleration due to
gravity (9.8 meters per second per second) times t²
Half the acceleration due to gravity is 4.9 m/sec²
So we can express height h at time t as:
h = 125 + 5.2t - 4.9t² When the package reaches the
ground, the value of h = zero, so we have the quadratic
equation: -4.9t² + 5.2t + 125 = 0
Solve for a positive value of t and we get 5.6 seconds.

Answer 3

This equation
y = Vo*t + (1/2)*a*t^2
will solve it but you will end up having to solve a quadratic equation.

These 2 equations
V^2 = Vo^2 + 2*g*y
y = (1/2)*(Vo+V)*t
will do it without a quadratic equation.

With either approach, be careful of the sign (polarity) of the velocities and acceleration.

Answer 4

s=V0*t - (1/2)*9.8*t² where s = 125m and V0 = 5.2 m/s

HTH

Doug

Answer 5

The vertical dist. to be covered= 125 m
Acceleration (due to gravity) = 9.8 m/Sec²
Initial velocity..............................= - 5.2 m/s(opp.Direction)

s = ut + ½at²
In this context,
Height = ut + ½gt²
125 = 5.2 t + ½ x 9.8 x t²
125 = 5.2 t + 4.9 t²
4.9t² + 5.2t -125 = 0
We can use the formula x = {- b ± √(b²-4ac)} / 2a
............................t..= { - 5.2 ± √{5.2² - 4 x 4.9 x (-125)} / 2 x 4.9
............................t.= { - 5.2 ± √27.04 + 2450} / 9.8
............................t..= { - 5.2 ± √2477.04} / 9.8
............................t.= { - 5.2 ± 49.77} / 9.8
............................t = 44.57 / 9.8 or - 54.97 / 9.8
The time will not be a -ve quantity.
............................t = 44.57 / 9.8 = 4.55 sec......Ans
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