A) \[\frac{1}{2}s\]
B) \[1\,s\]
C) \[\frac{3}{2}s\]
D) \[\frac{3}{4}s\]
Correct Answer: C
Solution :
Time taken by the ball to reach the plane \[t=\sqrt{\frac{2h}{g}}\] \[=\sqrt{\frac{2\times 4.9}{9.8}}=1\,s\] Height attained by the ball after 1st bounce \[h={{e}^{2}}h\] \[={{\left( \frac{3}{4} \right)}^{2}}\times 4.9\] \[\frac{1}{2}gt{{}^{2}}=\frac{9}{16}\times 4.9\] \[t=\sqrt{\frac{2\times 9\times 4.9}{16\times g}}\] \[=\sqrt{\frac{18\times 4.9}{16\times 9.8}}=\frac{3}{4}s\] So, the ball will strike second time after \[=2\times \frac{3}{4}=\frac{3}{2}s\]
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