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From the top of a building …

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From the top of a building 39.2m, a ball is thrown vertically upward with a velocity 9.8m–¹. Find the time when the ball will hit the ground.

Posted by Arunpreet Kaur 6 years, 6 months ago

CBSE > Class 11 > Physics

1 answers

Thakur Nobody 6 years, 6 months ago

Given u =9.8 a=-9.8 S=? Applying v=u+at 0=9.8+t×-9.8 -9.8=-9.8t T=1ses Time till it goes up =1sec Now V*2=u*2+2as 0*2=9.8*2 +2×-9.8×s 0=96.04-19.6s S=96.04/19.6 S=4.9m Hieght of building =39.2m Total hieght the ball reached=39.2+4.9=44.1 Now applying s=ut+1/2gt*2 44.1=0×t + 1/2×9.8×t*2 44.1=4.9×t*2 T*2=44.1/4.9 T*2=9 T=3sec

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