The height of a helicopter above the ground is given by h = 2.80t3, where h is in meters and t is in seconds. At t = 1.55 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? ...?
h = 2.80t^3
h = 2.80 (1.55)^3 = 10.43
10.43 = 1/2 gt^2
10.43 x 2/9.8 = t^2
t = √2.12
= 1.45
Hope this helps
Answer:
The time is 4.692 sec.
Explanation:
Given that,
Height \(h = 2.80t^3\)
Time t = 1.55 s
We know that,
The rate of change of height is the velocity.
So, the velocity is at t = 1.55 s
\(\dfrac{dh}{dt}= v = 3\times2.80\times(1.55)^2\)
\(v=20.181\ m/s\)
The velocity is upward with respect to the ground
We need to calculate the distance above the releasing point
Using equation of motion
\(v^2=u^2-2gs\)
Put the value into the formula
\(s=\dfrac{v^2}{2g}\)
\(s=\dfrac{20.181^2}{2\times9.8}\)
\(s=20.77\ m\)
The height of the helicopter releases a small mailbag
\(h=2.80\times(1.55)^3\)
\(h = 10.43\ m\)
We need to calculate the time
Using equation of motion
\(s=ut-\dfrac{1}{2}gt^2+h\)
Put the value into the formula
\(0=20.77\times t-\dfrac{1}{2}\times9.8\times t^2+10.43\)
\(t=-0.454,4.692\)
On neglecting negative value of time
Hence, The time is 4.692 sec.