A particle is projected vertically upwards with velocity $40 m / s$ . Find displacement and distance travelled by the particle in $6 s$
Take $g = 10 m / s^{2}$

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Solution

Time which it will reach its maximum is initial velocity/g $= 4 s e c .$
now the distance moved by ball in remaining $2 s e c$ is
$S = (\frac{1}{2}) g t^{2}$
$= (\frac{1}{2}) 10^{4} = 20 m .$
maximum height $= u t - (\frac{1}{2}) g t^{2} = 80 m$
therefore distance $100 m$ and displacement $= 80 - 20 = 60 m .$

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