A ball is dropped from a height of 90m on a floor. At each collision with the floor, the ball loses one tenth of its speed. plot the speed-time graph of its motion between t= 0 to 12 sec.
Answers
Explanation:
u=
Ball is drooped from a height s=90 mInitial velocity of the ball, u=0Acceleration, a=g=9.8m/s 2 Final Velocity of the ball =VFrom second equation of motion,S=ut+ 2/1at
1at ⇒90=0+ 2/1 ×9.8t
1 ×9.8t ⇒t= 18.38 =4.29s
1 ×9.8t ⇒t= 18.38 =4.29s
1 ×9.8t ⇒t= 18.38 =4.29s
1 ×9.8t ⇒t= 18.38 =4.29s From firstequationofmotion,V=u+at=0+9.8×4.29=42.04m/s
=0+9.8×4.29=42.04m/s
=0+9.8×4.29=42.04m/s Rebound velocity of the ball, ur= 10/9V = 10/9×42.04=37.84 m/s
9×42.04=37.84 m/sTime taken by the ball to reach maximum height is obtained with the help of first equation of motion as,
9×42.04=37.84 m/sTime taken by the ball to reach maximum height is obtained with the help of first equation of motion as,V=37.84+(−9.8)t ⇒t = −9.8−37.84 =3.86s
Total time taken by the ball =t+t =4.29+3.86=8.15 sAs the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time.The velocity with which the ball rebounds from the floor = 10/9×37.84 =34.05m/s
9×37.84 =34.05m/sTotal time taken by the ball for second rebound=8.15+3.86=12.01 s
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