You drop a 590g basketball out of your hand from rest. The basketball falls down a distance of 1.8m before striking the ground. Solve for the velocity that the basketball strikes the ground with. (assume this is an elastic collision) As the basketball strikes the ground, you hold your hand at 1.2m from the ground. A human hand is 400g. When the basketball bounces back up and colloids with your still hand, you allow your hand to move up with the basketball so they have the same final velocity. If 16% of the initial energy is lost in this collision. find the final height that your hand and the basketball are moved up to.

Answer rating: 100% (QA)

First lets calculate the velocity of the basketball when it strikes the ground We can do this using one of the equations of motion which is Vf2 Vi2 2ad where Vf is the final velocity Vi is the initial velocity which is zero in this case since the ball was dropped from rest a is acceleration gravity in this case which is 98 ms2 d is distance or height from which the ball was dropped which is 18 m Plugging in the values into the equation we get Vf2 0 2 98 18 Vf sqrt2 98 18 Vf 60 ms So the velocity of the basketball when it hits the ground is 60 ms Next when the ball bounces up and collides with your hand some of the balls energy is lost due to the collision Lets calculate this loss of energy If 16 of the initial kinetic energy is lost then only 84 100 16 of the energy is retained This means the basketball and the hand will move up with 84 of the basketballs original kinetic energy Kinetic energy can be calculated using the formula KE 12 m v2 where KE is the kinetic energy m is the mass v is the velocity Now lets find out the combined kinetic energy of the basketball after it bounces up and collides with your hand KEaftercoll 084 KEbeforecoll The basketballs kinetic energy before the collision can be calculated as KEbeforecoll 12 m v2 12 059 kg 60 ms2 1062 J So the combined kinetic energy of the basketball and your hand after the collision would be KEaftercoll 084 1062 892 J Since after the collision the basketball and the hand will move together having the same final velocity their combined kinetic energy would be KEaftercoll 12 massbasketball masshand vfinal2 Setting this equation equal to the kinetic energy after collision we found above and solving for vfinal we get 892 12 059 040 vfinal2 vfinal sqrt892 2 059 040 sqrt1784 099 425 ms Finally to find the height they moved up to we use the equation of motion again Vf2 Vi2 2ad Solving for d which represents the final height h h Vf2 Vi2 2a ... View the full answer