Answer to Question #217285 in Astronomy | Astrophysics for yarris

Question

What is the instantaneous velocity (in m/s) of a coconut falling vertically from a coconut tree after 2 2,0s?

Answer 1

Explanations & Calculations

• When a coconut falls from a coconut tree, it starts its motion from rest as it hangs in the tree at rest.
• Then the initial velocity of it is zero. $\small (u= 0\,ms^{-1})$
• And just after it loses its contact with the tree, it falls towards the ground under the gravitationl pull at the gravitational acceleration. So its acceleration is $\small a = g=9.8 ms^{-2}$.
• Since the acceleration is constant throughout the motion, the 4 equations of motion can be applied appropriately.
• When the time is given, and the above mentioned other parameters are known, the most suitable equation to calculate the instantaneous velocity turns out to be $\small v=u+at.$
• Then using this we get the result as follows,

$\qquad\qquad \begin{aligned} \small v&=\small u+gt\\ &=\small 0+9.8\,ms^{-2}\times2.0\,s\\ &=\small \bold{19.6\, ms^{-1}} \end{aligned}$

• The given value for the time is not clear, 22.0 or 2.0, anyway that cannot be 22.0s because then the coconut tree should be very tall or the coconut has reached the land & has come to rest.