9. GRAVITATION
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$19.6\, m$ ऊँची एक मीनार की चोटी से एक पत्थर छोड़ा जाता है। पृथ्वी पर पहुँचने से पहले इसका अंतिम वेग ज्ञात कीजिए।

A

$19.6\, ms^{-1}$

B

$15.8\, ms^{-1}$

C

$12.7\, ms^{-1}$

D

$13.5\, ms^{-1}$

Solution

According to the equation of motion under gravity $v^2 -u^2 = 2gs$

Where,

$u =$ Initial velocity of the stone $= 0\, m/s$

$v =$  Final velocity of the stone

$s =$ Height of the stone $= 19.6\, m$

$g =$ Acceleration due to gravity $= 9.8\, ms^{-2}$

$\therefore $ $v^2 – 0^2 = 2 \times 9.8 \times 19.6 $

$v^2 = 2 \times 9.8 \times 19.6 = (19.6)^2$

$v = 19.6 \,ms^{-1}$

Hence, the velocity of the stone just before touching the ground is $19.6\, ms^{-1}$

Standard 9
Science

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