A ball is thrown vertically downwards and the top of a building leaving the throws head with a velocity of 10 metre per second what is the velocity after falling for 2 seconds and how far it will fall into second and what will be its velocity after falling 10 m

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Solution

Dear Student
$v = u + g t v = 10 + 10 * 2 = 30 m / s e c v^{2} - u^{2} = 2 g S 30 \times 30 - 10 \times 10 = 20 \times S 800 = 20 S S = 40 m W h e n S = 10 m v^{2} - 100 = 20 \times 10 v^{2} = 300 v = 17.32 m / s e c R E G A R D S$

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A ball is thrown vertically downwards and the top of a building leaving the throws head with a velocity of 10 metre per second what is the velocity after falling for 2 seconds and how far it will fall into second and what will be its velocity after falling 10 m

Dear Student v=u+gtv=10+10*2=30m/secv2-u2=2gS30×30-10×10=20×S800=20SS=40mWhen S=10mv2-100=20×10v2=300v=17.32m/secREGARDS

Dear Student
$v = u + g t v = 10 + 10 * 2 = 30 m / s e c v^{2} - u^{2} = 2 g S 30 \times 30 - 10 \times 10 = 20 \times S 800 = 20 S S = 40 m W h e n S = 10 m v^{2} - 100 = 20 \times 10 v^{2} = 300 v = 17.32 m / s e c R E G A R D S$