Question

1 dimensional motion problems

Answer 1

A body falling freely under the action of gravity passess 2 points 40 m apart vertically in one second.Find from what height above the upper point it began to fall

Answer 2

help

Answer 3

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Answer 4

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Answer 5

\[\Delta s=40m\quad,\quad\Delta t=1s\]\[s_1=\frac{1}{2}gt^2\]\[s_2=s_1+\Delta s=\frac{1}{2}g(t+\Delta t)^2\]\[\Delta s=\frac{1}{2}g(t+\Delta t)^2-\frac{1}{2}gt^2=\frac{1}{2}g(2t+\Delta t)\Delta t\]\[t=\frac{1}{2}\left(\frac{2\Delta s}{g\Delta t}-\Delta t\right)=3.5s\]\[s_1=\frac{1}{2}gt^2=61.25m\quad(g=10m/s^2)\]

Answer 6

x = x0 - v0t - (gt^2)/2
x-x0 = 40
40 = -v0t -(gt^2)/2
it happened in 1 second so :
40 = -v0 - 5
v0 = -45 m/s.

now we need to calculate the time that it took it to reach 45 m/s from 0

v = -at
-45 = -10t
t= 4.5 sec

now the only thing to do is find the distance that it fell in 4.5 seconds with zero speed to begin with

x=(gt^2)/2
x = (10 * (4.5)^2 )/ 2 = 101.25 m
so 101.25 m above the upper point