Asecond baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. the throw is made with an initial speed of 18.0m/s at an angle of 37.5 above the horizontal. a) what is the horizontal component of the ball's velocity just before it is caught? b) how long is the ball in the air? . can someone explain to me how this works not just the answer i'm somewhat confused.

 (a) 

Taking the angle of the pitch, 37.5°, and the particle's initial velocity, 18.0 ms^-1, we get: 

18.0*cos37.5 = v_x = 14.28 ms^-1, the projectile's horizontal component. 

(b) 
To much the same end do we derive the vertical component: 

18.0*sin37.5 = v_y = 10.96 ms^-1 

Which we then divide by acceleration, a_y, to derive the time till maximal displacement, 

10.96/9.8 = 1.12 s 

Finally, doubling this value should yield the particle's total time with r_y > 0 

2.24 s

I hope my answer has come to your help. Thank you for posting your question here in .

 Energy used by television in 20 seconds = 3000 J

Explanation:

  We have expression for power, P = Work / Time

  Here we have power of television = 150 Watts

  Time of working of television = 20 seconds.

  We need to find energy it uses in 20 seconds, that is we need to find work done by electricity in 20 seconds.

  Work done = Power x Time = 150 x 20 = 3000 J

  Energy used by television in 20 seconds = 3000 J