## anonymous one year ago A 0.145 kg baseball is thrown horizontally at 15 m/s from a platform, so that it starts at 5.2 m above the ground. How much has its kinetic energy increased when it reaches the ground? 16 J , 24 J , 0 J , or 7.4 J ? :/

1. Michele_Laino

the initial kinetic energy is: $KE = \frac{1}{2}m{v^2} = \frac{1}{2} \times 0.145 \times {15^2} = ...Joules$

2. anonymous

so we get 16.3125? so our solution is 16 J?

3. Michele_Laino

that result is right! Nevertheless it is not our final answer!

4. anonymous

ohh okay! what is netx?

5. Michele_Laino

we have to consider that situation: |dw:1433103218745:dw| our ball, in free falling, has gained a vertical velocity

6. anonymous

yes:)

7. Michele_Laino

now, we have to compute that vertical speed

8. anonymous

ok! how do we do taht?

9. Michele_Laino

such vertical speed is given by the so called formula of Torricelli, which is: $v = \sqrt {2gh} = \sqrt {2 \times 9.81 \times 5.2} = ...m/\sec$

10. anonymous

10.10006 ? what happens now? :O

11. Michele_Laino

ok! Now we have this situation: |dw:1433103512624:dw|

12. Michele_Laino

namely the final velocity has the subsequent magnitude: ${v_f} = \sqrt {{{10}^2} + {{15}^2}} = ...m/\sec$

13. anonymous

18.0277?

14. Michele_Laino

ok! So the final kinetic energy is: $KE = \frac{1}{2}m{v^2} = \frac{1}{2} \times 0.145 \times {18^2} = ...Joules$

15. anonymous

okay! we get 23.49 so about 24 J is our solution?

16. Michele_Laino

no, since that the requested energy change is: $\Delta KE = 23.49 - 16 = ...Joules$

17. anonymous

ohh okay so we get 7.49 so about 7.4 J is our answer? :O

18. Michele_Laino

yes! that's right!

19. anonymous

yay!! thanks@!!

20. Michele_Laino

:)