Vertical Motion under Gravity Q)

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SUMMARY

The problem involves calculating the initial height (h) from which a stone is thrown vertically upwards with an initial speed of 16 m/s and hits the ground after 4 seconds. Using the kinematic equation for vertical motion, the displacement equation is set as -h = 16t - 4.9t². Substituting t = 4 seconds into the equation allows for the determination of h, leading to the conclusion that h equals 16(4) - 4.9(4)², which simplifies to h = 64 - 78.4, resulting in h = -14.4 meters, indicating the stone was thrown from a height of 14.4 meters above the ground.

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Homework Statement


Apologies for so many posts, however this is the last question.

A stone his thrown vertically upwards with speed 16m/s from point h meters above the ground. The stone hits the ground 4s later.
Find the value of h.

The Attempt at a Solution


I really haven't a clue on how to start this question, I haven't really encoutered this before, and would end up doing my own sledgehammer way.
But otherwise, can someone people give hints on how to solve this, I know all the kinematic equations, its just this situation confuses me cause of h.

Thanks in advance.
Lav.
 
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delta y = (voy)t - 4.9t^2

when the stone hits the ground, its vertical displacement will be the opposite of its initial height ...

delta y = -h

-h = 16t - 4.9t^2

sub in t = 4 ... h = ?
 

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