The height of a helicopter above the ground is given by h = 3.25t^3, where h is in meters and t is in seconds. At t = 2.25 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

chiodaiokaC

chiodaiokaC

Answered question

2022-11-26

Hello, Plainmath
Help me to solve this
The height of a helicopter above the ground is given by h = 3.25 t 3 , where h is in meters and t is in seconds. At t = 2.25 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Answer & Explanation

Gunnar Molina

Gunnar Molina

Beginner2022-11-27Added 8 answers

Answer:
In 2.748 sec the mailbag reached the ground
Solution:
We have given height from the ground h = 3.25 t 3
At t =2.25 sec helicopter releases a small mailbag so at t = 2.25 sec height from the ground h = 3.25 t 3 = 3.25 × 2.25 3 = 37.01 m
When the mail box is drooped its initial velocity would zero so u = 0 m/sec
Acceleration due to gravity g = 9.8 m / s e c 2
According to third law of motion h = u t + 1 2 g t 2
37.01 = 0 × t + 1 2 × 9.8 × t 2 t 2 = 7.553 t = 2.748 s e c

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