What is the speed of an apple dropped from a tree after $1.5$ second? What distance will it cover during this time? Take $g = 10 m / s^{2}$ .

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Solution

Given,
$u = 0, t = 1.5 s$

Speed of apple after 1.5 sec
From equation of motion-
$v = g t = 10 \times 1.5$
$v = 15 m / s$

distance $s = u t + \frac{1}{2} g t^{2}$
$= 0 + \frac{1}{2} \times 10 \times \frac{9}{4} = \frac{45}{4}$
$s = 11.25 m$

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What is the speed of an apple dropped from a tree after 1.5 second? What distance will it cover during this time? Take g=10m/s2.

Given,u=0,t=1.5sSpeed of apple after 1.5 secFrom equation of motion-v=gt=10×1.5v=15m/sdistance s=ut+12gt2 =0+12×10×94=454 s=11.25 m

What is the speed of an apple dropped from a tree after $1.5$ second? What distance will it cover during this time? Take $g = 10 m / s^{2}$ .

Given,
$u = 0, t = 1.5 s$

Speed of apple after 1.5 sec
From equation of motion-
$v = g t = 10 \times 1.5$
$v = 15 m / s$

distance $s = u t + \frac{1}{2} g t^{2}$
$= 0 + \frac{1}{2} \times 10 \times \frac{9}{4} = \frac{45}{4}$
$s = 11.25 m$