Answer

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Apne doubts clear karein ab Whatsapp par bhi. Try it now.

CLICK HERE

Watch 1000+ concepts & tricky questions explained!

21.5 K+

1.1 K+

Text Solution

Solution :

Taking vertical downward motion of ball from a height 90 m <br> We have <br> `u=0, a=10m//s^(2), S=90m, t= ?, v=?` <br> `t=sqrt((25)/a)=sqrt((2 times 90)/10)=3sqrt(25)=4.242` <br> `V=sqrt(2as)=sqrt(2 times 10 times 30)=30sqrt(2)m//s` <br> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/VIK_QB_PHY_XI_C03_E06_025_S01.png" width="80%"> <br> Rebound velocity of ball, <br> `u^(1)=9/10v=9/10 times 30sqrt(2)=27sqrt(2)m//g` <br> Time to reach the highest point is <br> `t^(1)=u^(1)/a=(27sqrt(2))/10=2.7sqrt(2)=3.81 S` <br> Total time `=t+t^(1)=4.24+3.81=8.05 S` <br> The ball will take further 3.1S to fall back to floor, where its velocity before striking the floor = `2.7sqrt(2)m//s`. <br> Time to reach the highest point is, <br> `t^(1)=u^(1)/a=(27sqrt(2))/(10)=2.7sqrt(2)=3.81S` <br> Total time `=t+t^(1)=4.24+3.81=8.05S` <br> The ball will take further 3.81 S fall back to floor, where its velocity before striking the floor `=2.7sqrt(2)m//s` <br> Velocity of ball after striking the floor <br> `=9/10times 27sqrt(2)=24.3sqrt(2)m//s`. <br> Total time elapsed before upward motion of ball. <br> `=8.05+3.81=11.86 S` <br> Thus the speed - time graph of this motion will be shown in fig.