Question

A helicopter is ascending vertically with a speed of 5.27m/s . At a height of 121m 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 initial speed equals the helicopter's.]

I filled in everything we have and I used: Y=y0 + Vy0t-(1/2)gt^2
--> I put: -9.8t^2+5.27t+121=0
And I did the quadratic formula and got t=3.7930,-3.2552. And used 3.79 s. However, the computer (this is online hw) says I am wrong! Help!

Answer 1

Take downward direction to be positive.
So, Initial velocity, \(u\)=-5.27m/s
Displacement, S=121m
Put \(S=ut+\frac{1}{2}gt^2\)
you will get your answer

Answer 2

so you're answer is 5.53 sec, right ?

Answer 3

what I did to get that answer is ;
[which may be wrong ;) ]

Given:
u=5.27m/s
h=121m (which may later called h1)

when helicopter dropped the package , in cannot falls directly, it must attain another height (like shown in the picture)


here;

h1 be the given height and
t2 the time for it
h2 be the height it attains when it is droppes and
t1 the time for that height

what we have to find out is Total time
which will be;
\[T = {t1} + {t2}\]

for t1:

u=5.27m/s
v=0
g=-9.8m/s^2

\[v=u+at1\]
\[{t1}=0.537s\]

for t2:

first we find h2;
\[h2=ut1 +1/2 (-g) t1^2\]
\[h2=1.419m\]

total height (H) = h1 + h2 = 121+1.419
H=122.419

now , we use H for t2
\[H=ut2+1/2g t2^2\]
122.419= 0*t2 +1/2 9.8t2^2
t2= 4.99
now;
T= t1 +t2= 0.537+4.99
\[T=5.53s\]

Answer 4

Thank you so much! You have no idea how much I appreciate your help! The problem makes alot more sense now! I went to see a tutor for this problem, but he did not give me the formula that you did! So thank you so much again. :)

Answer 5

@supercrazy92 bhai, your answer is correct but the solution could be simpler. Did you try using the formula i mentioned above. It will give you correct answer in a few steps!

Answer 6

lol, I didn't try that , thinking too for a shorter solution..

thanks for reminding me @ujjwal bhai ;)