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একজন লোক 49 মি/সেঃ বেগে একটি বল খাড়া উপরের দিকে নিক্ষেপ করে। বলটি কত সময় শূণ্যে অবস্থান করবে?
9.8 সে.
98 সে.
10 সে.
100 সে.
ANSWER : 3
Descrption
<p>উপরে ওঠার ক্ষেত্রে, </p><p>v=u-gt</p><p>=> 0 = 49 - 9.8 t<sub>1</sub></p><p>=> t<sub>1</sub> = 5 sec</p><p>নিচে নামার ক্ষেত্রে,</p><p>v=u+gt</p><p>=> 49 = 0 + 9.8 t<sub>2</sub></p><p>=> t<sub>2</sub> = 5 sec</p><p>so, T = t<sub>1</sub> +t<sub>2</sub> </p><p>T= 5 + 5 = 10 sec (Ans)</p>
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