Respuesta :
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
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 [tex]h = 2.80t^3[/tex]
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
[tex]\dfrac{dh}{dt}= v = 3\times2.80\times(1.55)^2[/tex]
[tex]v=20.181\ m/s[/tex]
The velocity is upward with respect to the ground
We need to calculate the distance above the releasing point
Using equation of motion
[tex]v^2=u^2-2gs[/tex]
Put the value into the formula
[tex]s=\dfrac{v^2}{2g}[/tex]
[tex]s=\dfrac{20.181^2}{2\times9.8}[/tex]
[tex]s=20.77\ m[/tex]
The height of the helicopter releases a small mailbag
[tex]h=2.80\times(1.55)^3[/tex]
[tex]h = 10.43\ m[/tex]
We need to calculate the time
Using equation of motion
[tex]s=ut-\dfrac{1}{2}gt^2+h[/tex]
Put the value into the formula
[tex]0=20.77\times t-\dfrac{1}{2}\times9.8\times t^2+10.43[/tex]
[tex]t=-0.454,4.692[/tex]
On neglecting negative value of time
Hence, The time is 4.692 sec.
