The height of a helicopter above the ground is given by h = 3.45 t 3 , where h i

Question

The height of a helicopter above the ground is given by h = 3.45t3, where h is in meters and t is in seconds. At t = 2.50 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?
s

————————————————————————

Three forces acting on an object are given by

and

The object experiences an acceleration of magnitude 4.00 m/s2.

(a) What is the direction of the acceleration?
  ° (counterclockwise from the +x-axis)

(b) What is the mass of the object?
kg

(c) If the object is initially at rest, what is its speed after 20.0 s?
  m/s

(d) What are the velocity components of the object after 20.0 s? (Let the velocity be denoted by

m/s

Explanation / Answer

Here ,

h = 3.45 *t^3

v = dh/dt = 3.45 * 3 * t^2

v = 10.35 * t^2

at t = 2.50 s

height of mailbag , h = 3.45 * 2.5^3

h = 53.9 m

velocity of mailbag , u = 10.32 * 2.5^2

u = 64.7 m/s

Now , let the time taken for the mailbag to reach is t

Using second equation of motion

d = u*t + 0.5 at^2

-53.9 = 64.7 * t - 0.5 * 9.8 * t^2

solving for t

t = 14 s

the time taken for the mailbag is 14 s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.